Clearly, $P$ is bounded, for it lies within $[0, 1].$ Each $E_n$ is composed of the union of $2^n$ closed intervals, and the union of finitely many closed intervals is also closed. $P$ is then the intersection of infinitely many closed intervals, which is again closed. Therefore, $P$ is closed and bounded, and by heine-borel, is compact.

$\square$

Suppose $E_n$ has $[\alpha, \beta]$ as an interval and thus $\alpha, \beta \in E_n.$ Then, by definition, $E_{n+1}$ will contain $[\alpha, \frac{\beta - \alpha}{3}]$ and $\frac{2(\beta - \alpha)}{3}, \beta]$ as intervals, so $\alpha, \beta \in E_{n+1}.$ Note that $E_0$ has $[0, 1]$ as an interval. By induction, all $E_n$ contain $0$ and $1$, and therefore so does their intersection $P,$ and $P$ is nonempty.

$\square$

Suppose, for the sake of contradiction, that some segment $(\alpha, \beta)$ \subset $P$ and let $L = \beta - \alpha.$ Pick some $n \in \mathbb{N}$ such that $1/3^n < L.$ Now, $E_n$ is the union of $2^n$ intervals of length $1/3^n,$ and since $(\alpha, \beta) \subset P,$ it must be the case that $(\alpha, \beta)$ is a subset of some interval of length $1/3^n.$ However, this can't be the case, since $L > 1/3^n,$ by construction. Therefore, our provision assumption is incorrect, and $P$ contains no segment.

$\square$

Let $x \in P.$ Let $r > 0,$ and pick $n \in \mathbb{N}$ to be large enough that $1/3^n < r.$ Then, $x$ lies in one of the $2^n$ intervals of length $1/3^n$ in $E_n;$ call it $I_n.$ The endpoints of $I_n$ are also in $P$ (see The Cantor set is not empty.,) and at least one of them is not $x.$ Since the endpoints are contained in a neighborhood of $x$ with radius $r > 0,$ all neighborhoods of $x$ are limit points of $P,$ and therefore $P$ is perfect.

$\square$