I have come up with a branched covering, but it necessarily has two branch points. From that I'm assuming that it can't be done, possibly related to the hairy ball theorem, but I don't know how to prove it either way.
The branched covering is two spheres which both have a great circle segment between two branch points along which they are connected. So if you're traveling on the first sphere and cross that segment, you are now traveling on the second sphere and vice-versa. Of course, at the branch points there is a discontinuity.
Is there a simple way to demonstrate that there is no such manifold? Or is there some construction which avoids the branch points?
I apologize for any misuse of terminology. Corrections welcome.
Because the sphere $S^{2}$ is simply-connected, it admits no connected, non-trivial covering. (If it matters, the trouble isn't related to the hairy ball theorem: The $3$-sphere admits continuous, nowhere-vanishing vector fields, but admits no connected, non-trivial covering, again because $S^{3}$ is simply-connected.)