This doesn't seem true to me. If then the open interval contains only .
Is there a flaw in this counterexample I'm not seeing?
In this case, I would try to construct a limit point.
First, can be contained in a closed interval . Split in half. One of the parts must contain infinitely many points from . Take that one and call it . Then, split in half. One of those parts must contain infinitely many points from , so take that one and call it . Continue this procedure indefinitely to receive a sequence of closed intervals such that whose sizes are , respectively.
Now, by the Nested Interval theorem, contains a single point, and that point is the limit point we are looking for.
Merely note that since is bounded we have that is bounded and thus compact. Thus, we have that there is an infinite subset of and by compactness (or the Bolzano-Weierstrass property to be exact) this sequence must have a limit piont. In any space (if this doesn't mean anything ignore it and replace it with ) a point is a limit point if and only if every neighborhood contains infintely many points of the set.