Uncountable Open Ordinal Space is not Separable
Let $\Omega$ denote the first uncountable ordinal.
Let $\hointr 0 \Omega$ denote the open ordinal space on $\Omega$.
Then $\hointr 0 \Omega$ is not a separable space.
Let $\sigma$ be the supremum of $H$.
Thus $\sigma$ strictly precedes $\Omega$.
The closed interval $\closedint 0 \sigma$ is such that $H \subseteq \closedint 0 \sigma$.
Thus the closure of $H$ does not equal $\hointr 0 \Omega$.
Thus $H$ is not everywhere dense in $\hointr 0 \Omega$.
Hence, by definition, $\hointr 0 \Omega$ is not separable.