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 and $\tau$ is the discrete topology.

Necessary Condition
Suppose that $\left({S, \tau}\right)$ is hereditarily compact.

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

Then $H$ is compact by hypothesis.

Since $\left({S, \tau}\right)$ is Hausdorff, we find that $H$ is closed in $\left({S, \tau}\right)$.

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

Thus $\tau$ is discrete.

Since $S \subset S$ is a subspace of $\left({S, \tau}\right)$, we find that $\left({S, \tau}\right)$ is compact.

Suppose $\left\vert{S}\right\vert = \infty$.

Then $\left({\left\{ {s}\right\} }\right)_{s \in S}$ is an open cover for $S$ that has no finite subcover.

Thus $\left\vert{S}\right\vert < \infty$.

Sufficient Condition
Let $S$ be finite with the discrete topology.

Since $S$ is finite, every open cover for $S$ is already finite, thus has a finite subcover.

Since $S$ 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.