How do I prove that every metric space that is sequentially compact and separable is compact? I can't seem to use either hypotheses.
I'll agree with your conclusion, but I don't know what "since we can map a dense subset into a countable subcover of A" means.X is sequentially compact and separable, and hence if an open cover A does not have a finite subcover, it has a countable subcover since we can map a dense subset into a countable subcover of A.
Why not? For example, if X is the set of real numbers and your cover is (n-1,n+1), take x_i=1 for the index corresponding to (-1,1) and x_i=0 for all other terms and you have a convergent sequence. I think you'll need your sets to be disjoint, and then pick an element in each set. I'm not 100% sure though. Sorry, I'm really busy now, but I'll look over this more carefully later tonight. If you need help sooner, maybe someone else has some ideas...Take an element that is in X\Ai for i=1 to infty and form a sequence, that sequence has a convergent subsequence that cannot converge to anything in the subcover, and hence outside X.