Suppose $K$ is a finite set in metric space $X$ and that $\{G_{\alpha }\},\alpha \in A$ is an open cover of $K.$ Since $K$ is finite, we can enumerate its points as $\{k_1,\dots ,k_n\},$ for some $n\ge 0.$ Then, for each $i=1,\dots ,n$ (there are none when $n=0$,) pick an $\alpha (i)\in A$ with $x_i\in G_{\alpha }(i).$ Define the index set $$A_0=\{\alpha (1),\dots ,\alpha (n)\}\subset A.$$ Because $n$ is finite, $A_0$ is finite as well, and $$K=\{k_1,\dots ,k_n\}\subset \bigcup _{\alpha \in A_0}{G}_{\alpha },$$ so $\{G_{\alpha }\},\alpha \in A_0$ is a finite sub-cover of the original open cover. Therefore, every open cover of $K$ has a finite sub-cover, and $K$ is compact.

Suppose $K$ is compact relative to $X$ and that $\{V_{\alpha }\}$ is an open cover of $K$ relative to $Y,$ such that $K\subset \bigcup _{\alpha }{V}_{\alpha }.$ We need to show that a finite subset of $\{V_{\alpha }\}$ covers $K.$ Because $\{V_{\alpha }\}$ is open relative to $Y$ and $Y\subset X$ (see Suppose $Y\subset X.$ A subset $E$...) there are sets $G_{\alpha },$ open relative to $X,$ such that $V_a=Y\cap G_{\alpha },$ for each $\alpha .$ Now, since $K$ is compact relative to $X,$ we have $$\begin{array}{cc}K\subset G_{{\alpha }_1}\cup \cdots \cup G_{{\alpha }_n},& \text{(a)}\end{array}$$ for some finite set of indices $a_1,\dots ,a_n.$ Now, since $K\subset Y,$ $$\begin{array}{cc}K\subset V_{{\alpha }_1}\cup \cdots \cup V_{{\alpha }_n},& \text{(b)}\end{array}$$ which shows $K$ is compact relative to $Y.$

Conversely, suppose $K$ is compact relative to $Y$ and let $\{G_{\alpha }\}$ be a cover of $K$ open relative to $X.$ We need to show there is a finite subset of $\{G_{\alpha }\}$ that covers $K.$ Let $V_{\alpha }=Y\cap G_{\alpha },$ for each $\alpha .$ Then (b) will hold for some set of indicies, ${\alpha }_1,\dots ,{\alpha }_n,$ and since each $V_{\alpha }\subset G_{\alpha },$ (a) is implied by (b) and we've shown $K$ is compact relative to $X.$

Suppose $K$ is compact relative to metric space $X.$ Let $p\in K^c.$ For each $q\in K,$ we can define $r_q=\frac{1}{2}d(p,q),$ and let $P_q$ and $Q_q$ be neighborhoods of radius $r_q$ around $p$ and $q,$ respectively. Note that $P_q$ and $Q_q$ are disjoint, because we defined their radii to be half the distance between them, and they are open.) Now, since $K$ is compact, we can pick a finite number of points in $K,$ $q_1,\dots ,q_n,$ such that $K\subset Q_{q_1}\cup \cdots \cup Q_{q_n}=Q$ ($Q$ is a finite subcover of $K.$) Using the same set of points as reference, let $P_{q_1}\cap \cdots \cap P_{q_n}=P.$ Note that $P\cap Q=\{\},$ since each $P_q$ is disjoint with its paired $Q_q$ (to be in $P$, a point must be in all $P_q,$ but any point in $Q$ is not in at least one $P_q.$) $P$ is open, since it is the intersection of finitely many open sets (see (a) - For any collection $\{G_a\}$ of...) and obviously contains $p,$ since each $P_q$ contains $p$. Therefore, $p$ has a neigborhood $P$ that is disjoint with $K$ (since $P\cap K\subset P\cap Q=\{\}$), and is therefore an interior point of $K^c.$ It follows that $K^c$ is open, and that $K$ is closed.

Suppose $F\subset K\subset X,$ with $F$ closed relative to $X,$ and $K$ compact. Let $\{V_{\alpha }\}$ be an open cover of $F.$ Since $F^c$ is open relative to $X$ (see A set $E$ is open iff its...), if we add it to $\{V_{\alpha }\},$ we obtain an open cover of $K;$ let's call it $\mathrm{\Omega }.$ Since $K$ is compact, we can obtain a finite subcover of $K$ by discarding all but a finite number of sets from $\mathrm{\Omega };$ let's call it $\mathrm{\Phi }.$ Since $F\subset K,$ $\mathrm{\Phi }$ is also a finite subcover of $F,$ and therefore $F$ is compact.

Let $G_{\alpha }=K_{\alpha }^c$ for each $\alpha ,$ and note that since $K_{\alpha }$ is compact and therefore closed, $G_{\alpha }$ is open. Then, fix a member $K_1$ of $\{K_{\alpha }\}.$ Assume, for contradiction's sake, that no point of $K_1$ is in all $K_{\alpha },$ that is, that $\bigcap K_{\alpha }=\mathrm{\varnothing }.$ Then, any point $x\in K_1$ is in some $K_{\alpha }^c=G_{\alpha },$ so $\{G_{\alpha }\}$ forms an open cover of $K_1.$ Since $K_1$ is compact, some finite subset $G_{{\alpha }_1},\dots ,G_{{\alpha }_n}$ of $\{G_{\alpha }\}$ forms a finite subcover of $K_1$ such that $K_1\subset G_{{\alpha }_1}\cup \cdots \cup G_{{\alpha }_n}={(K_{{\alpha }_1}\cap \cdots \cap K_{{\alpha }_n})}^c$ (by De Morgan's.) Therefore $K_1\cap K_{{\alpha }_1}\cap \cdots \cap K_{{\alpha }_n}=\mathrm{\varnothing }.$ This is an empty intersection of a finite subcollection of $\{K_{\alpha }\},$ which contradicts our hypothesis that all finite intersections are nonempty. Therefore, our assumption that no point in $K_1$ is in all $K_{\alpha }$ is incorrect, and some point in $K_1$ is in all $K_{\alpha },$ and therefore $\bigcap K_{\alpha }$ is not empty.

Assume, for the sake of contradiction, that no point in $K$ is a limit point of $E.$ Then any point $q$ in $K$ has a neighborhood with at most one point in $E;$ $q,$ if $q\in E.$ Since $E$ is infinite, an infinite number of these singleton neighborhoods would be required to cover it, and therefore to cover $K,$ since $E\subset K.$ But, this contradicts our hypothesis that $K$ is compact. Therefore, our provisional assumption must be false, and $K$ must contain a limit point of $E.$

Let $I_n=[a_n,b_n],$ and let $E$ be the set of all $a_n.$ Then, $E$ is nonempty, because even if $a_n=b_n$ for all $n,$ it at least contains a single point. It is also bounded above by $b_1,$ since any $b_n$ is in $[a_1,b_1].$ Let $x=supE.$ Let $m$ and $n$ be positive integers and we have that $$a_n\le a_{m+n}\le b_{m+n}\le b_m,$$ so that $x\le b_m$ for each $m.$ Since $a_m\le x,$ we have that $a_m\le x\le b_m,$ that is, $x\in I_m$ for all $m=1,2,3,\dots ,$ so $x\in \bigcap _{i=1}^n{I}_n$ and thus $\bigcap _{i=1}^n{I}_n$ is not empty.

Let $I$ be a $k$-cell, consisting of all points $\mathbf{x}=(x_1,\dots ,x_k)$ such that $a_j\le x_j\le b_j,1\le j\le k.$ Let $$\delta =\sqrt{\sum _{j=1}^k(b_j-a_j{)}^2},$$ i.e., the maximum distance between any two points in $I$ (the diagonal). Then for any points $\mathbf{x},\mathbf{y}\in I,|\mathbf{x}-\mathbf{y}|\le \delta .$

Suppose, for the sake of contradiction, that there is an open cover $\{G_{\alpha }\}$ of $I$ that contains no finite subcover of $I.$ Now, let $c_j=(a_j+b_j)/2,$ i.e. $c_j$ is the midpoint of $[a_j,b_j].$ We can subdivide $I$ into $2^k$ $k$-cells $Q_i,$ determined by the intervals $[a_j,c_j]$ and $[c_j,b_j].$ At least one $Q_i,$ call it $I_1,$ cannot be covered by any finite subcollection of $\{G_{\alpha }\},$ or else $I$ would have a finite subcover in $\{G_{\alpha }\}.$ We then can subdivide $I_1$ and so on, obtaining a sequence $\{I_n\}$ with the following properties:

(a) $I_{n+1}\subset I_n$ (each $k$-cell in the sequence is nested in the previous.)

(b) $I_n$ is not covered by any finite subset of $\{G_{\alpha }\}.$

(c) If $\mathbf{x},\mathbf{y}\in I_n,$ then $|\mathbf{x}-\mathbf{y}|\le 2^{-n}\delta .$

From (a) and If $I_n$ is a sequence of intervals..., there is some point ${\mathbf{x}}^{\ast }$ that is in every $I_n.$ For some $\alpha ,x^{\ast }\in G_{\alpha },$ since $\{G_{\alpha }\}$ is a cover of all of $I.$ $G_{\alpha }$ is open, so for some implies that $y\in G_{\alpha },$ that is, $\mathbf{x}$ has a neighborhood that lies entirely within $G_{\alpha }.$ If we make $n$ big enough, we have that so by (c), $I_n\subset G_{\alpha },$ that is, $I_n$ is entirely covered by $G_{\alpha }.$ But, this contradicts (b), so our provisional assumption is incorrect, and $\{G_{\alpha }\}$ must have a finite subcover that covers $I,$ and therefore $I$ is compact.

If (a) holds, then $E\subset I$ for some $k$-cell $I,$ and (b) follows from the facts that every $k$-cell is compact and closed subsets of compact sets are compact. Then, (c) follows from the fact that any infinite subset $K$ of a compact set $E$ has a limit point in $E$. To complete the cycle of implication, we must now show that (c) implies (a).

Assume, for the sake of contradiction, that $E$ is not bounded. Then, $E$ must contain an indexed set of points $\{x_n\},n=1,2,3,\dots$ where each $x_n$ must satisfy $\{x_n\}$ is obviously infinite, but we will show it has no limit points in $E.$ Let $p\in E.$ Then for some positive integer $N,$ Since $N$ is finite, there can only be finitely many points $\{q_{\alpha }\}$ in $\{x_n\}$ with If $\{q_{\alpha }\}$ is empty, then $p$ is obviously not a limit point of $\{x_n\}.$ Otherwise, let $r=min\{d(p,q_{\alpha })\}.$ Then and let $N_r\{p\}$ be a neighborhood of $p.$ Since no point in $\{x_n\},$ other than perhaps $p,$ lies in $\{x_n\},$ $p$ is clearly not a limit point of $\{x_n\}.$ Thus, our provisional assumption is invalid and (c) implies $E$ is bounded.

To show that (c) implies $E$ is closed, assume for the sake of contradiction that $E$ is not closed. Then, there is a point $x_0\in R^k$ which is a limit point of $E$ but is not in $E.$ We will construct an infinite subset of $E$ and show that it has no limit point in $E.$ For $n=1,2,3,\dots ,$ let the point $x_n$ be some point in $E$ such that let $\{x_n\}$ be the set of such points. $\{x_n\}$ is certainly infinite, because $|x_n-x_0|$ will eventually be bigger than $1/n$ for some $n$ if we keep reusing the same $x_n$ infinitely many times. Now, $\{x_n\}$ has $x_0$ as a limit point, and we will show it is its only limit point in $R^k.$ Assume $y\in R^k,y\ne x_0.$ Then, via the triangle inequality, for all but finitely many $n,$ and thus $y$ is not a limit point of $\{x_n\}$ because its its neighborhoods do not contain infinitely many points of $\{x_n\}$. Thus, $\{x_n\}$ has no limit point in $E,$ which contradicts (c), and therefore our provisional assumption that $E$ is not closed is incorrect, and (c) implies that $E$ is closed.

Suppose $E$ is a bounded infinite subset of $R^k.$ Then, it is a subset of a $k$-cell $I\subset R^k,$ and because every $k$-cell is compact, $I$ is compact. Since infinite subsets of a compact set $K$ have a limit point in $K$, $E$ has a limit point in $I$ and therefore in $R^k.$