Let (X, d) be a metric space.
Show that any finite subset is compact.
While , what h2osprey said was undeniably correct, I think the finer points got lost.
To be compact means that every open cover has a finite subcover. Thus, to show that a set is compact one way is to take an arbitrary open cover and derive a finite subcover. So, the concept is we let be some open cover of . Next, we let be arbitrary, and since covers there exists some open set, which we will call , in such that it contains . So, for each we take some corresponding and construct a new class . This is clearly a subcover, and since there is at most elements of it must be a finite subcover.
The two finer points that you should definitely answer yourself is why can I choose this and why is there at most elements in