Maybe you need the Banach space because of compacity.
I don't know if it is correct, but consider the following:
In a metric space, a set is closed and bounded it is compact.
So choosing a sequence of and the corresponding neighborhoods in which
the function f restricted to these neighborhoods is Lipschitz with constants , you can always
pick a finite sequence to still cover the whole bounded (and closed) set.
If the set is open, wasn't there some theorem that one can, for a continuous function, extend it to the boundary continuously?
Well, with closed bounded sets it works fine - find some argument for open ones. ;-)
Then space doesn't need to be Banach though, it only has to be a metric space.