Real Numbers are Uncountably Infinite/Cantor's Second Proof

Proof
By definition, a perfect set is a set $X$ such that every point $x \in X$ is the limit of a sequence of points of $X$ distinct from $x$.

From Real Numbers form Perfect Set, $\R$ is perfect.

Therefore it is sufficient to show that a perfect subset of $X \subseteq \R^k$ is uncountable.

We prove the equivalent result that every sequence $\sequence {x_k}_{k \mathop \in \N} \subseteq X$ omits at least one point in $X$.

Let $y_1 \in X$.

Let $B_1:= \map {\BB_r} {y_1}$ be a closed ball centred at $y_1$.

Consider a closed ball:
 * $B_{n - 1} \supseteq B_n := \map {\BB_{\map \delta n} } {y_n}$

such that:
 * $\map \delta n \le \dfrac {\map \delta {n - 1} } 2$
 * $y_n \in X$
 * $x_n \notin B_n$

Note that $\map \delta 1 = r$.

We can satisfy the condition $x_n \notin B_n$ because $X$ is perfect, so every ball centred at a point of $X$ contains infinitely many points of $X$.

Since $\map \delta n \le \dfrac r {2^{n - 1} }$, $y_n$ is Cauchy.

Therefore from Perfect Set is Closed, we may let $\displaystyle Y = \lim_{n \mathop \to \infty} y_n \in X$.

For $n \in \N$ we have:
 * $\set {y_m: m > n} \subseteq B_n$

so $Y \in B_n$.

But by construction:
 * $\forall n \in \N: x_n \notin B_n$

Therefore:
 * $\forall n \in \N: Y \ne x_n$

and the proof is complete.