This is a lot harder than I initially thought. The idea behind the following is that this is obvious if

is bounded, and if it isn't we can "erase" the parts that are far enough away from

to be ignored.

So, we know that

is compact and thus totally bounded (specifically bounded). Thus,

for some constant

. Let

. This is a compact subspace of

and so

is compact (since the intersection of a compact set and a closed set is compact). Now, we next claim that

, but I leave this for you to prove. Thus, we must merely show that

for some

. To do this we note that since

are compact that

is compact by Tychonoff's theorem. Also, it is clear that

is continuous and so it assumes it's minimum on

and the conclusion follows.