= Solution
Take $(x_0,y_0)$ outside the <graph of a function>, so $y_0\ne f(x_0)$. Since $Y$ is a <Hausdorff space>, choose disjoint open neighborhoods $V$ of $y_0$ and $W$ of $f(x_0)$. <Continuity> makes $U=f^{-1}(W)$ an open neighborhood of $x_0$. The product neighborhood $U\times V$ contains $(x_0,y_0)$ and cannot meet the graph, because every $f(x)$ with $x\in U$ belongs to $W$. The complement is therefore open in the <product topology>, proving
$$
\boxed{G_f\text{ is closed in }X\times Y}.
$$
This is the <closed graph of a map into a Hausdorff space> property; no separation assumption on $X$ is required.
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