Compact Hausdorff Topology is Minimal Hausdorff

Theorem
Let $T = \left({X, \tau}\right)$ be a Hausdorff space which is compact.

Then $\tau$ is the minimal subset of $\mathcal P \left({T}\right)$ which is a Hausdorff space.

Proof
Suppose there exists a topology $\tau'$ on $X$ such that $\tau' \subseteq \tau$.

Then the identity mapping $I_X: \left({X, \tau}\right) \to \left({X, \tau'}\right)$ would be continuous.

So from Closed Subspace of Compact Space is Compact if $A$ is closed in $\left({X, \tau}\right)$ it must be compact.

Thus $I_X \left({A}\right)$ must also be compact.

If $\left({X, \tau'}\right)$ were a Hausdorff space then from Compact Subspace of Hausdorff Space is Closed $I_X \left({A}\right)$ would be closed.

Thus $I_X$ would be a closed mapping and so $\tau \subseteq \tau'$.

So no topology which is strictly coarser than $\tau$ can be Hausdorff.