Finite Topological Space is Compact

Theorem
Let $T = \left({S, \vartheta}\right)$ be a topological space where $S$ is a finite set.

Then $T$ is countably compact.

Proof
Let $\mathcal V$ be a countable open cover of $T$.

Then $\forall x \in S: \exists \mathcal V_x \subseteq \mathcal V: \forall V_x: x \in V_x$.

We set up a choice function to select one element of $V_x$.

Thus we create a finite subcover of $\mathcal V$

Hence the result by definition of countably compact.