Compact First-Countable Space is Sequentially Compact

Theorem
Let $\left({X, \vartheta}\right)$ be a compact metric space. Then, every sequence in $X$ has a convergent subsequence.

That is: a compact metric space is sequentially compact.

Proof
Consider a sequence $\left \langle {x_n} \right \rangle_{n \in \N}$ in $X$.

Reasoning by contradiction, assume that it has no convergent subsequence. That is, that for every point $x \in X$, there is no subsequence of $\left \langle {x_n} \right \rangle$ that converges to $x$.

Then, for every $x \in X$ one can find an open set $U_x \subseteq X$ which contains $x$ and contains only a finite number of terms of the sequence $\left \langle {x_n} \right \rangle$.

(The contrary of this is to say that there is a subsequence of $\left \langle {x_n} \right \rangle$ which converges to $x$).

The family $\mathcal U = \left\{{U_x : x \in X}\right\}$ is an open cover of $X$.

As $X$ is compact, we can extract from $\mathcal U$ a finite subcover, so all of $X$ is covered a finite number of the $U_x$.

But each of these only contains a finite number of the $x_i$, and this implies that there are only a finite number of terms in the sequence $\left \langle {x_i} \right \rangle$, which is absurd.