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 $\mathcal C$ of $T$ must have $S$ in it.

So $\left\{{S}\right\}$ will be a subcover of $\mathcal C$, whatever $\mathcal C$ may be.

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

Hence the result, by definition of compact space.