Uncountable Open Ordinal Space is not Separable

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

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

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

Proof
Because $\Omega$ is the first uncountable ordinal, any ordinal which strictly precedes $\Omega$ is countable.

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

Let $\sigma$ be the supremum of $H$.

As $H$ by definition strictly precedes $\Omega$, $H$ itself is countable.

Thus $\sigma$ strictly precedes $\Omega$.

The closed interval $\left[{0 \,.\,.\, \sigma}\right]$ is such that $H \subseteq \left[{0 \,.\,.\, \sigma}\right]$.

By definition of the closure $H^-$ of $H$ as the smallest closed set of $\left[{0 \,.\,.\, \Omega}\right)$ containing $H$, it follows that $H^- \subseteq \left[{0 \,.\,.\, \sigma}\right]$.

Therefore, there exists an open interval $\left({\sigma \,.\,.\, \Omega}\right)$ in the complement of $H^-$ in $\left[{0 \,.\,.\, \Omega}\right)$.

Thus the closure of $H$ 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.