Uncountable Open Ordinal Space is not Lindelöf

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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 Lindelöf space.


Aiming for a contradiction, suppose $\hointr 0 \Omega$ is a Lindelöf space.

From Ordinal Space is Completely Normal, $\hointr 0 \Omega$ is a completely normal.

From Sequence of Implications of Separation Axioms, $\hointr 0 \Omega$ is a $T_3$ space.

From Lindelöf $T_3$ Space is Paracompact, it follows that $\hointr 0 \Omega$ is a paracompact space.

But this contradicts the fact that from Uncountable Open Ordinal Space is not Paracompact, $\hointr 0 \Omega$ is not a paracompact space.

Hence the result by Proof by Contradiction.