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
From Power Set of Finite Set is Finite, the power set of $S$ is finite.

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

Let $\mathcal V$ be an open cover of $S$.

By definition $\mathcal V \subseteq \tau$ and so is also a finite set.

From Set is Subset of Itself, $\mathcal V \subseteq \mathcal V$.

Thus by definition $\mathcal V$ is a finite subcover of $\mathcal V$.

The result follows by definition of compact.