Hausdorff Space is Hereditarily Compact iff Finite

Theorem
Let $\left({S, \tau}\right)$ be a Hausdorff space.

Then $\left({S, \tau}\right)$ is hereditarily compact $S$ is finite.

Necessary Condition
Let $\left({S, \tau}\right)$ be hereditarily compact.

That is, every subspace of $\left({S, \tau}\right)$ is compact.

Let $H \subset S$ be a subspace of $\left({S, \tau}\right)$.

From Compact Subspace of Hausdorff Space is Closed, $H$ is closed.

Thus for all $H \subset S$, we find that $H$ is closed in $S$.

It follows by definition of discrete topology that $\tau$ is discrete.

But since $S$ is compact and Discrete Space is Compact iff Finite, $S$ is also finite.

Sufficient Condition
Let $S$ be finite.

From $T_2$ (Hausdorff) Space is $T_1$ Space, $\left({S, \tau}\right)$ is a $T_1$ (Fréchet) space.

From Finite T1 Space is Discrete, it follows that $\left({S, \tau}\right)$ has the discrete topology.

As $S$ is finite, every open cover for $S$ is therefore finite.

Thus $S$ has a finite subcover.

As $S$ is finite, $H \subset S$ is also finite, and every open cover for $H$ is also finite.

Thus every open cover for $Y$ has a finite subcover.

Thus $\left({S, \tau}\right)$ is hereditarily compact.

Also see

 * Indiscrete Space is Hereditarily Compact, an example of a non-Hausdorff, infinite, non-discrete hereditarily compact space.