Sequentially Compact Space is Countably Compact

Theorem
Every sequentially compact space is a countably compact space.

Proof
Let $T = \left({X, \vartheta}\right)$ be a sequentially compact space.

By definition:
 * $T$ is a countably compact space if every countable open cover of $T$ has a finite subcover.

By definition of sequentially compact space, every infinite sequence in $T$ has a subsequence which converges to a point in $T$.

By Equivalent Definitions of Countably Compact, it suffices to show that every infinite sequence in $X$ has an accumulation point in $X$.

The result follows from Limit Point of Sequence is Accumulation Point.