Compact Hausdorff Topology is Minimal Hausdorff

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

Then $\vartheta$ is minimally Hausdorff.

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

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

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

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

If $\left({X, \vartheta'}\right)$ 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 $\vartheta' \subseteq \vartheta'$

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