Theorem $\hspace{5 pt}$ Let $P$ be a nonempty perfect set in $\mathbb{R}^k$. Then $P$ is uncountable.
Proof $\hspace{5 pt}$ Since $P$ has limit points, $P$ must be infinite. Suppose $P$ is countable, and denote the points of $P$ by $\mathbf{x_1}, \mathbf{x_2}, \mathbf{x_3}, \ldots$. We shall construct a sequence $\{V_n\}$ of neighborhoods as follows. Let $V_1$ be any neighborhood of $\mathbf{x_1}$. If $V_1$ consists of all $\mathbf{y} \in \mathbb{R}^k$ such that $|\mathbf{y} - \mathbf{x_1}| < r$, the closure $\overline{V_1}$ of $V_1$ is the set of all $\mathbf{y} \in \mathbb{R}^k$ such that $|\mathbf{y} - \mathbf{x_1}| \leq r$. Suppose $V_n$ has been constructed, so that $V_n \cap P$ is not empty. Since every point of $P$ is a limit point of $P$, there is a neighborhood $V_{n+1}$ such that (i) $\overline{V_{n+1}} \subset V_n$, (ii) $\mathbf{x_n} \notin \overline{V_{n+1}}$, (iii) $V_{n+1} \cap P$ is not empty. By (iii), $V_{n+1}$ satisfies our induction hypothesis, and the construction can proceed. Put $K_n = \overline{V_n} \cap P$. Since $\overline{V_n}$ is closed and bounded, $\overline{V_n}$ is compact.
I've understood things up until this point.
Since $\mathbf{x_n} \notin K_{n+1}$, no point of $P$ lies in $\bigcap_1^\infty K_n$.
1) I understand that $\{\mathbf{x_n}\}$ is a subset of $P = \{\mathbf{x_1}, \mathbf{x_2}, \mathbf{x_3}, \ldots\}$. If this subset was equal to $P$ then I can see how no point of $P$ would lie in the intersection. But I don't think this is true (since $P$ would have to be bounded in order for $V_1 \supset P$ so that $\mathbf{x_n}\in V_n \subset V_{n-1} \ldots \subset V_1$).
What's the actual reason why $\mathbf{x_n} \notin K_{n+1}\implies\forall p \in P, p \notin \bigcap_1^\infty K_n$ ?
Update I've been Trying to understand Quang Hoang's answer for 1)
"Every element of $P$ is of the form $x_n$ for some $n$, hence doesn't lie in $K_n$."
But doesn't $\{\mathbf{x_n}\}$ only includes elements within the first neighborhood? (Since each $\mathbf{x_n}\in V_n$ for a particular $V_{n}\subset V_{n-1}$, surely $\{\mathbf{x_n}\}$ can't contain every point in $P$?)
What am I missing?
Since $K_n \subset P$, this implies that $\bigcap_1^\infty K_n$ is empty. But each $K_n$ is nonempty, by (iii), and $K_n \supset K_{n+1}$, by (i); this contradicts the Corollary to Theorem 2.36.
Resolved 2) In order to apply Theorem 2.36, $K_n$ must be compact. How do we know that $K_n$ is compact? EDIT: For 2) I understand the answer given by Quang Hoang, thanks!