Compact First-Countable Space is Sequentially Compact

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

That is: a compact topological 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.