Uncountable Closed Ordinal Space is not Separable

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

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

Then $\left[{0 \,.\,.\, \Omega}\right]$ is not a separable space.

Proof
Let $H \subseteq \left[{0 \,.\,.\, \Omega}\right]$ be a countable subset of $\left[{0 \,.\,.\, \Omega}\right]$.

From Uncountable Open Ordinal Space is not Separable, there exists an open interval $\left({\sigma \,.\,.\, \Omega}\right)$ in the complement of $H^-$ in $\left[{0 \,.\,.\, \Omega}\right)$, and so also in $\left[{0 \,.\,.\, \Omega}\right]$.

Thus the closure of $H$ in $\left[{0 \,.\,.\, \Omega}\right]$ does not equal $\left[{0 \,.\,.\, \Omega}\right]$.

Thus $H$ is not everywhere dense in $\left[{0 \,.\,.\, \Omega}\right]$.

Hence, by definition, $\left[{0 \,.\,.\, \Omega}\right]$ is not separable.