The two statements are equivalent in a metric space:
1. Every infinite sequence has a convergent subsequence. In other words, there is an accumulation point for every infinite sequence.
2. The metric space is convergent.
Proving (2) from (1) follows the standard practice of deducing that any open set containing the accumulation point contains all but finite points in the sequence, and the remaining finite points can be covered by a finite number of open sets.
I used to wonder what if there are infinite accumulation points? Can there be infinite sequences, each with infinite subsequences converging to infinite separate points? The answer is yes. But we notice that using the above argument is not enough.
However, the problem soon gets resolved if we notice that the set of all limit points of a sequence (the limit points of all subsequences) is closed! See if you can manage to figure out why this necessarily produces an open covering. If you’re unable to, leave a comment. I will include the full solution.