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 countably compact first-countable topological space.

We need to show that $T$ is sequentially compact, i.e., that every infinite sequence in $S$ has a convergent subsequence.

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

By Infinite Sequence in Countably Compact Space has Accumulation Point, $\left \langle {s_n}\right \rangle$ has an accumulation point $p \in S$.

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 $p$.