Is a finite Hausdorff space necessarily discrete?
Yes. Better, it works for T1, too: T1 is the axiom that onepoint sets are closed. Then since the set is finite, the complement of any point is also closed; the point is open. That's the discrete topology. 


Just thought of another answer to this. Any topology on a finite set is compact. Any map from a discrete topology is continuous. Hence by the famous theorem on maps from compact spaces into Hausdorff spaces, the identity map on a finite space is a homeomorphism from the discrete topology to the given Hausdoff topology. A certain phrase involving sledgehammers and walnuts springs to mind, though. 


Yes. Let X be finite and Hausdorff. It is enough to show that every point x in X is open. For every point y different from x, there is an open neighborhood U_*y* of x not meeting y. The intersection of the U_*y*'s is open and equals {x}. 

