Page "Heine–Borel theorem" Paragraph 9

If S is compact but not closed, then it has an accumulation point a not in S. Consider a collection consisting of an open neighborhood N ( x ) for each x ∈ S, chosen small enough to not intersect some neighborhood V < sub > x </ sub > of a.

Then is an open cover of S, but any finite subcollection of has the form of C discussed previously, and thus cannot be an open subcover of S. This contradicts the compactness of S. Hence, every accumulation point of S is in S, so S is closed.