If "S" is compact but not closed, then it has a limit 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""x" 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 limit point of "S" is in "S", so "S" is closed.