I've been able to provide a fix for the proof given in the book. I'm fairly sure the author made a thinko. The proof can be fixed since it is enough to prove the theorem for dense in . For instance if we can prove that is closed in , then its also closed in .
So assuming that its easy to prove the statement I was having trouble with:
Let be a compact neighbourhood of (in the subspace topology ).
That is a neighbourhood of in means that there exists an open neighbourhood of such that .
Because is open it is easy to show that for any we have that . In particular we have .
Since is dense in this means that and so is a closed neighbourhood of in such that .