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?

The crux of the argument is that we make sure to collect all the points. List the elements $\{x_{1}, x_{2}, \cdots , x_{n}, \cdots\}$ and then take the open interval $U_1$. If $x_{2}$ is not in $U_{1}$, ignore it, as it will not show up in $V$. If $x_{2}$ is in $U_{1}$, we construct $U_{2}$. In this way, we go down the list of $x_{n}$, and we ignore them if they are not in current $U_{j}$ we are considering, and otherwise we construct $U_{j+1}$. When we build $V$, and prove it is nonempty, we have shown that we have either actively eliminated points or ignored those that could not have ended up in $V$. Thus we get a contradiction.
