Countably Compact First-Countable Space is Sequentially Compact

Theorem
Let $T = \left({X, \vartheta}\right)$ be a first-countable space.

Then $T$ is sequentially compact iff $T$ is countably compact.

Proof
Let $T = \left({X, \vartheta}\right)$ be a first-countable space.

We have that a Sequentially Compact Space is Countably Compact whether the space is first-countable or not.

So what we need to show is that if $T = \left({X, \vartheta}\right)$ is countably compact it follows that it is also sequentially compact.

So, let $T = \left({X, \vartheta}\right)$ be a countably compact.

Then every countable open cover of $X$ has a finite subcover.

We need to show that every infinite sequence in $X$ has a subsequence which converges to a point in $X$.

Let $\left \langle {s_n}\right \rangle$ be any sequence in $X$.

Let $p \in X$ be an accumulation point of $\left \langle {s_n}\right \rangle$.

As $T$ is first-countable, $p$ has a countable neighborhood basis, say:
 * $\left\{{V_n: V_1 \supseteq V_2 \supseteq V_3 \supseteq \cdots}\right\}$

Then a subsequence $\left \langle {s_{n_i}}\right \rangle$, where $s_{n_i} \in V_i$, converges to $p$.