Subset of Indiscrete Space is Compact and Sequentially Compact

Theorem
Let $T = \left({S, \left\{{\varnothing, S}\right\}}\right)$ be an indiscrete topological space.

Let $H \subseteq S$.

Then $H$ is compact in $T$ and sequentially compact in $T$.

Proof
The subspace $T_H = \left({H, \left\{{\varnothing, S \cap H}\right\}}\right)$ is trivially also an indiscrete space.

The only open cover of $T_H$ is $\left\{{H}\right\}$ itself.

The only subcover of $H$ is, trivially, also $\left\{{H}\right\}$, which is finite.

So $H$ is (trivially) compact in $T$.

From Convergent Sequences in Indiscrete Space, every sequence in $T$ converges to every point of $S$.

So every infinite sequence has a subsequence which converges to every point in $S$.

Hence $H$ is (trivially) sequentially compact in $T$.