This is good.

There is an easier way to go about this. Let . Then, for each you can form disjoint open balls which contain respectively. Note then that evidently is an open cover for and so by assumption has some finite subcover . Now tell me, why is a neighborhood of disjoint from ?now to show that A is closed too.

here is the part where i am stuck:

Let be a sequence in such that . I must show that . i tried to use contradiction method, that is, i assumed that there exists a sequence in with such that . now i must exhibit an open covering of which does NOT admit a finite subcovering in order to bring up a contradiction but i am not able to go further.