Well, I assume you meant since you can't compare complex numbers. Other than that, and assuming you're using open disks there, I think your general idea works.
Q: Let . Prove that is an open set.
A: Let by arbitrary. We need to find a such that the disk centered at , is completely contained in . So, let . Then for any we have that . Therefore, and we are done.
Is my choice of legit? It seams reasonable that disk with equal to the smaller of the two distance from the edge of the set will be in the set, since the neighborhood itself is open.