Let be an open set. Suppose that the map is a and is .
Prove that .
The crucial piece of information here is that the function is in fact uniformly continuous. I'm still working on figuring out what that entails.
I think I have it. Suppose . Then is not closed (since only closed and open sets are the empty and whole sets). In particular there exists a such that there exists a sequence with each . Then since the sequence is convergent, it is Cauchy, and since the function is uniformly continuous the sequence of images is Cauchy. Since is complete, there exists such that . Since the function was onto, there exists u such that .
Then since is continuous, implies . But so by uniqueness of limits . Since we assumed but , we have our contradiction!
Convergence is a stronger condition than Cauchy since to be convergent we must identify the element the sequence converges to, whereas with cauchy we only know the sequence elements get closer together.