Rudin 2.43 Every nonempty perfect set in $\mathbb{R}^k$ is uncountable

I am having trouble understanding Rudin's proof of this theorem, and I believe I have pinpointed the particular part of the proof that I am unable to follow. To begin with, here is the proof copied from Baby Rudin:

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 $x_1, x_2, x_3, \dots$. We shall construct a sequence $\{V_n\}$ of neighborhoods as follows.

Let $V_1$ be any neighborhood of $x_1$. 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 $|y - x_1| \leq r$.

Suppose $V_n$ has been constructed so that $V_n \cap P \neq \emptyset$. 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) $V_{n+1} \cap P \neq \emptyset$. 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 $x \notin K_{n+1}$ no point of $P$ lies in $\cap_1^\infty K_n$. Since $K_n \subset P$ this implies $\cap_1^\infty$ 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.

Okay so my confusion lies with this paragraph of the proof:

Suppose $V_n$ has been constructed so that $V_n \cap P \neq \emptyset$. 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) $V_{n+1} \cap P \neq \emptyset$. By (iii) $V_{n+1}$ satisfies our induction hypothesis and the construction can proceed.

My confusion is as follows:

In defining $V_1$, it is clear that $V_1$ is a neighborhood of the point $x_1$. Is $V_{n+1}$ a neighborhood of the point $x_{n+1}$? For example, is $V_2$ a neighborhood of the point $x_2$? I don't think this can be the case because there is no guarantee that there is a neighborhood of $x_2$ is a subset of an arbitrary neighborhood of $x_1$. After all if $x_1 \neq x_2$ and we can choose any neighborhood of $x_1$ in defining $V_1$, it is certainly possible $x_2 \notin V_1$. So if $V_{n+1}$ is not a neighborhood of $x_{n+1}$, what point is $V_{n+1}$ a neighborhood of?

• Possible duplicate of Problems in Theorem 2.43 of baby Rudin Feb 12, 2018 at 10:14
• @JoséCarlosSantos I don't think this is a duplicate: although both this and the linked question are about Theorem 2.43, this question asks if $x_{n+1} \in V_{n+1}$, while the linked question asks to show that $V_{n+1}$ exists.
– user88319
Feb 12, 2018 at 20:42
• @Strants You are right. I have retracted my closing vote. Feb 12, 2018 at 20:58

Note that nowhere in the proof is used the fact that $x_n$ should lie in $V_n$. The only point is that $V_n$ intersects $P$, so it is a neighbourhood of at least one point in $P$, no matter which one.
The fact that $x_1\in V_1$ may be confusing, but that's just to initiate the sequence.
Note: to construct $V_{n+1}$ simply select any open ball centered around a point in $V_n\cap P$ that is not $x_n$, with radius small enough that the closed ball lies entirely in $V_n$. All requirements are trivially satisfied.
• Perhaps it is presumed $x_n \in V_n$ for all n. Feb 12, 2018 at 10:19
• @WilliamElliot No it's not, you don't need that at all. All you need is for $V_n$ to not contain any of the $n$ first $x_n$, so that the intersection in the end is empty. Feb 12, 2018 at 10:21
It is not assumed that $$x_n \in V_n$$ (it may or may not be). Regardless of whether $$x_n \in V_n$$ or $$x_n \not \in V_n$$, we want to construct $$V_{n+1}$$ such that $$x_n \not \in V_{n+1}$$, and I do so here: Proof of Baby Rudin Theorem 2.43