Uncountable Open Ordinal Space is Countably Compact

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

Let $\hointr 0 \Omega$ denote the open ordinal space on $\Omega$.

Then $\hointr 0 \Omega$ is a countably compact space.

Proof
Let $\closedint 0 \Omega$ denote the closed ordinal space on $\Omega$.

From Uncountable Closed Ordinal Space is Countably Compact, $\closedint 0 \Omega$ is a countably compact space.

So every sequence in $\hointr 0 \Omega$ has an accumulation point in $\closedint 0 \Omega$.

But $\Omega$ cannot be an accumulation point of any sequence in $\closedint 0 \Omega$.

So every sequence in $\hointr 0 \Omega$ has an accumulation point in $\hointr 0 \Omega$.

This means that $\hointr 0 \Omega$ is countably compact.