Compact Hausdorff Topology is Maximally Compact

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

Then $\vartheta$ is maximally compact.

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

Consider the identity mapping $I_X: \left({X, \vartheta'}\right) \to \left({X, \vartheta}\right)$.

From Separation Properties Preserved in Subspace, $I_X$ would be a continuous bijection from a Hausdorff space to a compact Hausdorff space.

If $\left({X, \vartheta'}\right)$ is also compact, then $I_X$ must also be open, and so $\vartheta' \subseteq \vartheta$.

So no topology which is strictly finer than $\vartheta$ can be compact.