Countably Compact First-Countable Space is Sequentially Compact/Proof 2

Theorem
A countably compact first-countable topological space is also sequentially compact.

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

Let $T = \left({S, \tau}\right)$ be countably compact.

By definition, every countable open cover of $S$ has a finite subcover.

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

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

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

As $T$ is first-countable, $p$ has a countable local 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 $p$.