Modified Fort Space is Compact

Theorem
Let $T = \left({S, \tau_{a, b}}\right)$ be a modified Fort space.

Then $T$ is a compact space.

Proof
Let $\mathcal C$ be an open cover of $T$.

Then $\exists U \in \mathcal C: a \in U$.

Then by definition of modified Fort space, $U$ is cofinite.

That is, $S \setminus U$ is finite.

Then $S \setminus U$ is covered by a finite subcover of $\mathcal C$.

Hence, by definition, $T$ is compact.