I suggest the following idea.

Now f is a bijection, so we only need to check that, if is open in the domain (, d), then A is open in the range (, d).

Suppose not so. For some set A of (, d), such that is open but A is not open. Then we have some element y of A, such that every -neighborhood contains a point not in A. Construct a sequence {}, such that each is not an element of A, but d(, y)