Difficulty: 4

Suppose a set $A$ is sequentially closed. Let $p \in X$ and $(x_\alpha)_{\alpha < \lambda} \subseteq A$. Take $V$ a countable nbd of $p$. If $x_\alpha \to p$, then there is an ordinal $\mu$ for which $x_\alpha \in V$ for all $\alpha \ge \mu$. The set of ordinals $\mu \le \alpha < \lambda$ is isomorphic to some $\lambda'$. So we have $(y_\alpha)_{\alpha < \lambda'} \subseteq V$, still with $y_\alpha \to V$.

From a converging transfinite sequence within a countable set, we can extract a $\omega$-sequence that still converges to the same value, this was done in (T211). So we construct a sequence $(z_n)$ with $z_n \to p$, and so $p \in A$ as we wished to prove.

$X$ is a countable union of finite open sets. $X$ hereditarily connected, so this basis is linearly ordered and it’s possible to enumerate them as $\{B_n\}$ where $n < m \implies B_n \subset B_m$. Therefore, $B_n \nearrow X$ and $X$ is a union of finite sets.