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True. If there were only finitely many isolated points, delete all of them. The remainder would be a nonempty closed, countable, complete subspace, so part (v) would give it an isolated point. Its positive distance from the finite deleted set would make it isolated in the original space as well, a contradiction. Thus the stronger isolated points in a countable complete metric space result applies.

True. If a nonempty countable complete metric space had no isolated point, every singleton would be closed and nowhere dense. Their countable union would be the whole space, contradicting the Baire category theorem. This is the basic isolated points in a countable complete metric space argument.