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.

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.

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 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 by construction. Therefore, our provision assumption is incorrect, and $P$ contains no segment.

Let $x\in P.$ Let and pick $n\in \mathbb{N}$ to be large enough that 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 all neighborhoods of $x$ are limit points of $P,$ and therefore $P$ is perfect.