Assume $A$ is countably infinite, $E\subset A,$ and $|E|=\mathrm{\infty }.$ Arrange a sequence $\{x_n\}$ from the distinct elements of $A$. Let $n_1$ be the smallest positive integer such that $x_{n_1}\in E,$ and then pick $n_2,\dots ,n_{k-1}$ by assigning the next $n$ in the sequence the index of the left-most entry in $\{x_n\}$ that has not yet been picked. Then, let $n_k$ be the smallest integer greater than $n_{k-1}$ such that $x_{n_k}\in E.$ Now, $\{n_k\}$ is a sequence of strictly increasing positive integers giving us the indices of the first $k$ elements of $E$ in $\{x_n\}.$

Now, define as $f(k)=x_{k_n}$, which is a bijection between the positive integers and $E$, showing that $E$ is countable.

We can construct an infinite array where the rows are sequence constructed by the sets that make up the entries of $\{E_n\}.$ Then, we can create a single sequence from all the entries of the sets of $\{E_n\}$ by iterating over them in the following order:

sequence = []
for i in range(1, k):
    for j in range(0, i):
        n = i - j
        m = j + 1
        sequence.append(f"E_{n},{m}")

Which, for $k=5,$ yields

$$E_{1,1},E_{2,1},E_{1,2},E_{3,1},E_{2,2},E_{1,3},E_{4,1},E_{3,2},E_{2,3},E_{1,4}.$$

This sequence may contain duplicates, so some indices may need to be skipped in constructing a subset $T$ of the positive integers such that $TS,$ but we've now shown that $S$ is at most countable. To show $S$ is infinite and therefore countable, note that the infinite set $E_1$ is a subset of $S,$, and therefore $S$ is infinite and countable.

We will proceed using proof by induction. First, for the base case, note that $B_1$ is the set of $1$-tuples formed by elements of $A$, so $B_1=A$ and is thus countable. Now, for the inductive step, assume $B_{n-1}$ is countable $(n=2,3,4,\dots ).$ Then we have that

$$B_n=\{(b,a)|b\in B_{n-1},a\in A\}=\bigcup _{b\in B_{n-1}}(b\times A).$$

So, for any given $n-1$-tuple $b$, we form $n$-tuples by appending each element of $a$ to it, and so the set of pairs $(b,a)$ has the same cardinality as $A,$ and is thus countable. $B_n$ is thus the union of the countable set of countable sets (the set of sets formed by appending each element of $A$ to each element of $B_{n-1}$) and is therefore countable itself, by a theorem proved above. Therefore, by induction, every $B_n$ is countable.

Let $E$ be a countable subset of $A,$ and call the elements of $E$ $s_1,s_2,s_3,\dots .$ We will construct a new sequence $p$ in the following way:

$$p_n=\mathrm{\neg }{s}_{n_n},n=1,2,3,\dots$$

That is, the $n$th digit of $p$ will be the opposite of whatever the $n$th digit of $s_n$ is. So, $p$ differs from $s_1$ in the first digit, from $s_2$ in the second digit, $s_3$ in the third digit, and so on, so that it differs from all elements of $E,$ and therefore is not contained in $E.$ But, $p$ is definitely in $A$ since it its elements are the digits $0$ and $1.$ Therefore, $E$ is a proper subset of $A,$ so any countable subset of $A$ must be a proper subset of $A.$ But, $A$ can't be a proper subset of itself, and therefore $A$ must be uncountable.