Subspace of Finite Complement Topology is Compact

Theorem
Let $T = \left({S, \tau}\right)$ be a finite complement topology on an infinite set $S$.

Then every topological subspace of $T$, including $T$ itself, is a compact space.

Proof
Let $T_H = \left({H, \tau_H}\right)$ be a subspace of $T$.

Let $\mathcal C$ be an open cover of $T_H$.

Let $U \in \mathcal C$ be any set in $C$.

$U$ covers all but a finite number of points of $T_H$.

So for each of those points we pick an element of $\mathcal C$ which covers each of those points.

Hence we have a finite subcover of $T_H$.

So by definition $T_H$ is a compact space.