Compact First-Countable Space is Sequentially Compact

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

This is: a compact topological space is sequentially compact.

Proof
Consider a sequence $$\{x_n\}_{n \in \N}$$ in $$X$$. Reasoning by contradiction, assume that it has no convergent subsequence, this is, that for every point $$x \in X$$, there is no subsequence of $$\{x_n\}$$ that converges to $$x$$.

Then, for every $$x \in X$$ one can find an open set $$U_x$$ which contains $$x$$ and contains only a finite number of terms of the sequence $$\{x_n\}$$ (as the contrary of this is to say that there is a subsequence of $$\{x_n\}$$ which converges to $$x$$).

The family $$\{ U_x \mid x \in X \}$$ is an open cover of $$X$$. As $$X$$ is compact, we can extract from this 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 $$\{x_i\}$$, which is absurd.