I think there is a technicality that you need to be careful of here.
Let be the sequence . Then is a sequence in , but it is not a subsequence of .
Thus your statement following "by assumption" is not in fact by assumption, but must be proved.
I am working on this problem, and I am not 100% sure with my argument. Could someone help me verify it?
Let (X,d) be a metric space. Suppose that is a sequence in X such that it has no convergent subsequence. Claim: is closed in X.
By assumption, for every sequence in A, does not converge. Otherwise, has a convergent subsequence. This implies that A is closed since I'll show that a set A is closed in (X,d) iff for every sequence in A if converges then .
First direction: Suppose A is closed. Let be a convergent sequence in A. Assume , then . But X-A is open and implies that there exists N such that if n>N, then . Contradicts to our assumption that is a sequence in A.
Second direction: I'll show X-A is open. Let be a sequence in X converging to . It suffices to show that there exists N such that if n>N, then . Assume for only finitely many times, then is a sequence in A. By hypothesis, . But . Contradiction. So, X-A is open, ie A is closed.
Thanks a lot for your correction, DrSteve. I should have seen that.
Thanks Plato. I think I can prove that statement. Could you verify my argument for me? Since is a limit point of , every neighborhood of has nonempty intersection with . So, there exists . The sequence converges to p. But how can I be sure that this is a sequence of distinct points. Also, I don't see how proving this means I'm done with the original question. Could you elaborate on that please?