Problems in Theorem 2.43 of baby Rudin 
Theorem 2.43 Let $P$ be a nonempty perfect set in $\mathbb{R}^k$. Then $P$ is uncountable.
Proof  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 $y\in \mathbb{R}^k$ such that $|y−x_1|<r$, the closure $\overline{V_1}$ of $V_1$ is the set of all $y\in \mathbb{R}^k$ such that $|y−x_1|≤r$.
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) $x_n\notin \overline{V_{n+1}}$, (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. Since $\mathbf{x_{n}}\notin K_{n+1}$, no point of $P$ lies in $\cap_1^\infty K_n$. Since $K_{n}\subset P$, this implies that $\cap_1^\infty K_n$ 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.

I have not been able to understand third paragraph of Walter's proof. I would like to understand why the neighborhood $V_{n+1}$ exists with properties (i),(ii) and (iii), using only the previous definitions and theorems of the Book.

I explain why there is a neighborhood $V_n$ satisfying (i), (ii) and (iii). We know that $P$ is perfect. That is, every point in $P$ is a limit point of $P$. Especially, $x_n$ is a limit point of $P$ so every neighborhood $V$ of $x_n$ contains a point in $P$ differ from $x_n$. Let it say $y$.
You can take a neighborhood $W = B(y, \rho)$ of $x_{n+1}$ which is not contain $x_n$ and contained in $V_n$. If we take $ V_{n+1} = B(y,\rho/2) $, then $\overline{V_{n+1}} \subseteq W\subseteq V_n$. Since $x_n\notin W$, $\overline{V_{n+1}}$ does not contain $x_n$. Moreover, since $y\in P$ and $y\in V_{n+1}$, $V_{n+1}\cap P$ is not empty.
Let me know if you need more details or have another question.
