Help understanding proof of Theorem 2.43 in Baby Rudin 
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!
 A: *

*Every element of $P$ is of the form $x_n$ for some $n$, hence doesn't lie in $K_n$. 

*$K_n$ is a closed subset (intersection of two closed sets) of a compact set.

A: Ok, I've figured out the cause of my confusion:

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 $\mathbf{x_n}\in V_n \subset V_{n-1} \ldots \subset V_1$).

Let $V_n$ be the neighborhood of $\mathbf{y_n}$. Then $\mathbf{x_1} = \mathbf{y_1}$. I realise now that even if $\mathbf{x_n} \neq \mathbf{y_n}$, we can still construct $V_{n+1}$ so that (ii) is satisfied:
Since $\mathbf{x_n}$ is an arbitrary point, and there are infinitely many limit points of $P$ in $V_n$, we can choose $\mathbf{y_{n+1}}\in P, V_n$ such that $\mathbf{y_{n+1}} \neq \mathbf{y_{n}}$ and $\mathbf{y_{n+1}}\neq \mathbf{x_{n}}$.
Now since every point of $P$ is a limit point, we can pick a small enough $r>0$ such that $V_{n+1}$ satisfies (i)-(iii).
