This question arose when I was reading a book on topology. In fact, it is a book on locally compact groups, hence only briefly discusses the topic. And the question is that, while in the proposition of the theorem, it is stated that when a map $\phi$ maps a topological space $X$ to another $Y$, and when $X$ is Hausdorff, $Y$ is compact and the graph of $\phi$ is closed, then $\phi$ 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.
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.