Countable Closed Ordinal Space is Second-Countable

Theorem
Let $\Omega$ denote the first uncountable ordinal.

Let $\Gamma$ be a limit ordinal which strictly precedes $\Omega$.

Let $\left[{0 \,.\,.\, \Gamma}\right]$ denote the closed ordinal space on $\Gamma$.

Then $\left[{0 \,.\,.\, \Gamma}\right]$ is a second-countable space.

Proof
From Basis for Open Ordinal Topology, the set $\mathcal B$ of subsets of $\left[{0 \,.\,.\, \Gamma}\right]$ of the form:
 * $\left({\alpha \,.\,.\, \beta + 1}\right) = \left({\alpha \,.\,.\, \beta}\right] = \left\{ {x \in \left[{0 \,.\,.\, \Gamma}\right): \alpha < x < \beta + 1}\right\}$

for $\alpha, \beta \in \left[{0 \,.\,.\, \Gamma}\right)$, forms a basis for $\left[{0 \,.\,.\, \Gamma}\right]$.

As $\Gamma$ strictly precedes $\Omega$, there are a countable number of points of $\left[{0 \,.\,.\, \Gamma}\right]$.

Each point in $\left[{0 \,.\,.\, \Gamma}\right]$ has a countable basis.

From Countable Union of Countable Sets is Countable, it follows that $\left[{0 \,.\,.\, \Gamma}\right]$ has a countable basis.