Theorem A map $\phi$ maps a topological space $X$ to another $Y$, where $X$ 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 $C$.