Compact Subset of Compact Space is not necessarily Closed

Theorem
A closed subset of a compact space not necessarily closed.

Proof
Let $S$ be a set containing more than one element.

Let $\tau = \left\{ {S, \varnothing}\right\}$ be the indiscrete topology on $S$.

Let $x \in S$.

Let $H = S \setminus \left\{ {x}\right\}$.

Then $H$ is a proper subset of $S$.

Then from Subset of Indiscrete Space is Compact and Sequentially Compact, the subspace induced by $\tau$ on $H$ is a compact subspace of $\left({S, \tau}\right)$.

But from Open and Closed Sets in Indiscrete Topology, the only closed subsets of $\left({S, \tau}\right)$ are $S$ and $\varnothing$.

Thus $H$ is a compact subspace of $\left({S, \tau}\right)$ which is not closed in $S$.