Finite Topological Space is Compact/Proof 2

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

Then $T$ is compact.

Proof
Since $S$ is finite, its power set is finite.

Since the topology $\tau$ is by definition a collection of subsets of $S$, it follows that $\tau$ is finite as well.

Thus, if $\mathcal V$ is an open cover of $S$, then by definition it is a subset of $\tau$, and consequently it is already a finite subcover.

The result follows by definition of compact.