Theorem A map ϕ maps a topological space X to another Y, where X is Hausdorff, Y is compact and the graph of ϕ is closed. Then ϕ is continuous.
Is it reallly necessary to include the condition that X is Hausdorff? Since I see no reason, and I appear to have a proof without using the condition, I would like to know the answer.
A proof:
For any closed subspace C of Y, the pre-image D ought to be closed. For any element a in the complement of D, we can use the compactness to show that there is a finite number of open sets in Y such that the union of them covers the C, and then the corresponding open sets in X is an open neighborhood of a which has an empty intersection with D; so D is closed.
P.S. in the proof, those open sets are obtained by the condition that the graph is closed and that (a,c) is not in the graph for any c in C.