Real Numbers are Uncountably Infinite/Cantor's Second Proof

Theorem
The set of real numbers $\R$ is uncountably infinite.

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 $(x_k)_{k \in \N} \subseteq X$ omits at least one point in $X$.

Let $y_1 \in X$, and let $B_1 := \mathcal B_r(y_1)$ be some closed ball centred at $y_1$.

Given $B_{n-1}$ choose a closed ball $B_{n-1} \supseteq B_n := \mathcal B_{\delta(n)}(y_n)$ such that $\delta(n) \leq \delta(n-1)/2$ (note that $\delta(1) = r$), $y_n \in X$ and $x_n \notin B_n$.

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 $\displaystyle \delta(n) \leq \frac r{2^{n-1}}$, $y_n$ is Cauchy.

Therefore since a perfect set is necessarily closed we may let $\displaystyle Y = \lim_{n \to \infty} y_n \in X$.

For $n \in \N$ we have $\{ y_m : m > n\} \subseteq B_n$, so $Y \in B_n$.

But by construction for each $n \in \N$ $x_n \notin B_n$.

Therefore $Y \neq x_n$ for all $n \in \N$, and we are done.

Historical Note
This proof was first demonstrated by.