The idea is that for each point , there are disjoint (i.e. non-overlapping) neighborhoods and of and , respectively. (Notice that is fixed but can be any point in .) In order to ensure they are disjoint, we need to make sure they have a small enough radius. How small? Well if they are no more than half the distance between and , that will be small enough. So it doesn't really matter exactly how are defined as long as they are disjoint neighborhoods. If you like we can define them as you have done. But that's just one possible choice. All we need them to be are disjoint neighborhoods.

And then of course we have an open cover of . By compactness there is a finite subcover which is a finite set of the form for some . Remember that each is disjoint from each . So is disjoint from (verify this through the algebra of sets), and since we have a neighborhood of . So is an interior point, and since is arbitrary, all points in are interior---which is equivalent to saying that is open.

EDIT: You have several typos in your OP. For example the theorem should read "compact subsets of metric spaces areclosed." And your LaTeX code is off sometimes. Please note that not everyone can see through to what you mean.