Excluded Point Space is Compact/Proof 2

Proof
By definition of excluded point space, the only open set of $T$ which contains $p$ is $S$.

So any open cover $\CC$ of $T$ must have $S$ in it.

So $\set S$ will be a subcover of $\CC$, whatever $\CC$ may be.

And $\set S$ (having only one set in it) is trivially a finite cover of $T$.

Hence the result, by definition of compact topological space.