I figured out part b, but part a is confusing me.
If is convergent then all subsequences converge to the same limit. Thus, the forward direction follows.
To prove the backwards direction assume that [u]did not[/b] converge to . Then it means there is some so that for any we have for some . Thus, we can form a subsequence, with . But this subsequence cannot possible have a subsequence converging to . Contradiction.