Code borrowed from here.
Let $P$ be a non-empty perfect set in $\mathbb{R}^k$. Then $P$ is uncountable.
Here's the definition of a perfect set:
Let $(X,d)$ be a metric space, and let $P \subset X$. Then $P$ is perfect if it is closed (i.e. it contains all of its limit points) and every point of $P$ is also a limit point of $P$.
Now here's the proof Rudin gives:
Since $P$ has limit points, $P$ must be infinite. Suppose $P$ is countable, and denote the points of $P$ by $x_1, x_2, x_3, \ldots$. 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 $\vert y – x_1 \vert < r$, the closure $\overline{V_1}$ of $V_1$ is the set of all $y \in \mathbb{R}^k$ such that $\vert y – x_1 \vert \leq r$.
Supose $V_n$ 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) $x_n \not\in \overline{V_{n+1}}$, (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 $x_n \not\in K_{n+1}$, 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 non-empty, by (iii), and $K_n \supset K_{n+1}$, by (i); this contradicts the Corollary to Theorem 2.36.
Finally, here's Theorem 2.36:
If $\{K_\alpha\}$ is a collection of compact subsets of a metric space $X$ such that the intersection of every finite subcollection of $\{K_\alpha\}$ is nonempty, then $\cap K_\alpha$ is nonempty.
And, here's the Corollary to Theorem 2.36:
If $\{K_n\}$ is a sequence of nonempty compact sets such that $K_n \supset K_{n+1}$ ($n=1, 2, 3, \ldots$), then $\cap_1^\infty K_n$ is not empty.
I don't understand why it is always possible to construct $V_n$ with these properties. Does we choose random points and then denote them with numbers or we have to choose $x_n$ in some specific way ?
"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) $x_n \not\in \overline{V_{n+1}}$, (iii) $V_{n+1} \cap P$ is not empty." I don't understand this part why there is exist $V_{n+1}$ such that (i) $\overline{V_{n+1}} \subset V_n$ and what does the limit point have to do with that .
Best Answer
Without reading through the related questions and comments, let me answer your concrete questions in a concrete way.
Let's start, as in the book, with $x_1\in P$ and an arbitrary neighborhood $V_1$ of $x_1$. I will construct $V_2$ and then explain how a general induction would go.
Here comes in the limit point and perfect condition already. Since $P$ is perfect, $x_1$ is a limit point of $P$ (also called an accumulation point by many authors), so any neighborhood of $x_1$ contains a point of $P$ different from $x_1$. Therefore $V_1$, being a neighborhood of $x_1$, contains another point of $P$. Since $P$ is assumed to be $\{x_1,x_2,\dots\}$, we have an $x_{n_1}\in V_1$ different from $x_1$. For later use, we assume $n_1$ is the smallest such index (this exists by the well-ordering principle).
Since $x_{n_1}\neq x_1$, (by ${\mathbb R}^k$ being Hausdorff and regular, if you are comfortable with point-set topology), we have a neighborhood $V_2$ of $x_{n_1}$ such that (i) $\overline{V_2}\subset V_1$, (ii) $x_1\notin \overline{V_2}$ (you can concretely shrink the radius of $V_2$ to make this true), (iii) $V_2\cap P$ is not empty (since it contains $x_{n_1}$). That is the construction of $V_2$.
Now you can go totally general to talk about the general inductive construction like the book did. The whole thing feels a bit awkward since the condition for $V_3$ is that $x_2\notin \overline{V_3}$.
Let's proceed in a more direct way. By our assumption of $n_1$, the points $x_2,\dots,x_{n_1-1}\notin V_1$, so they are not in $\overline{V_2}$. We can choose $V_2=\dots=V_{n_1}$. I think requiring (i) to be $\overline{V_{n+1}}\subset \overline{V_n}$ would be totally sufficient for this proof.
Eventualy, we have $x_{n_1}\in V_{n_1}=V_2$, and we repeat what we did for $x_1\in V_1$ to get $V_{n_1+1}$, which again will be equal all the way to $V_{n_2}$.
In the end, we satisfy the following slightly modified conditions (i) $\overline{V_{n+1}}\subset \overline{V_{n}}$, (ii) $x_n\notin \overline{V_{n+1}}$, (iii) $V_n\cap P$ is not empty.
The proof would be completed as before by these conditions.
(I do want to personally remark that I always find Rudin's proofs hard to follow. They are never what I would think first. Now explaining this particular proof makes me feel that the author can definitely be more gentle toward the reader.)