Proof that a perfect set is uncountable There is something I don't understand about the proof that perfect sets are uncountable. The same proof is present in Rudin's Principles of Mathematical Analysis.
Do we assume that our construction of $U_n$ must contain all points of $S$? What if we are only collecting evenly-indexed points of $S$ ($x_{2n}$)? We would still get an infinitely countable subset of $S$, and the rest of $S$ can be used to provide points for $V$. What am I missing?

There is an alternative proof, using what is a consequence of Baire's Theorem:

THM Let $(M,d)$ be a complete metric space with no isolated points. Then $(M,d)$ is uncountable.

PROOF Assume $M$ is countable, and let $\{x_1,x_2,x_3,\dots\}$ be an enumeration of $M$. Since each singleton is closed, each $X_i=X\smallsetminus \{x_i\}$ is open for each $i$. Moreover, each of them is dense, since each point is an accumulation point of $X$. By Baire's Theorem, $\displaystyle\bigcap_{i\in\Bbb N} X_i$ must be dense, hence nonempty, but it is readily seen it is empty, which is absurd. $\blacktriangle$.

COROLLARY Let $(M,d)$ be complete, $P$ a perfect subset of $M$. Then $P$ is uncountable.

PROOF $(P,d\mid_P)$ is a complete metric space with no isolated points.

ADD It might be interesting to note that one can prove Baire's Theorem using a construction completely analogous to the proof suggested in the post. 

THM Let $(X,d)$ be complete, and let $\langle G_n\rangle$ be a sequence of open dense sets in $X$. Then $G=\displaystyle \bigcap_{n\in\Bbb N}G_n$ is dense.

PROOOF We can construct a sequence $\langle F_n\rangle$ of closed sets as follows. Let $x\in X$, and take $\epsilon >0$, set $B=B(x,\epsilon)$. Since $G_1$ is dense, there exists $x_1\in B\cap G_1$. Since both $B$ and $G_1$ are open, there exists a ball $B_1=B(x_1,r_1)$ such that $$\overline{B_1}\subseteq B\cap G_1$$
Since $G_2$ is open and dense, there is $x_2\in B_1\cap G_2$ and again an open ball $B_2=B(x_2,r_2)$ such that $\overline{B_2}\subseteq B_1\cap G_2$, but we ask now that $r_2\leq r_1/2$. We then successively take $r_{n+1}<\frac{r_n}2$. Inductively, we see we can construct a sequence of closed bounded sets $F_n=\overline{B_n}$ such that $$F_{n+1}\subseteq F_n\\ \operatorname{diam}D_n\to 0$$
Since $X$ is complete, there exists $\alpha\in \displaystyle\bigcap_{n\in\Bbb N}F_n$. But, by construction, we see that $\displaystyle\alpha\in \bigcap_{n\in\Bbb N}G_n\cap B(x,\epsilon)$
Thus $G$ is dense in $X$.$\blacktriangle.$
