Graph Criterion for Continuity into Compact Spaces (3.26.8)

James Munkres Topology, chapter 3.26, exercise 8:

Let $Y$ be a compact Hausdorff space. Then show $f : X→Y$ is continuous if and only if the graph of $f$ $$G_f = \{x×f(x)~|~x∈X\}$$ is closed in $X×Y$.

Proof: ($⇒$) Suppose $f$ is continuous. We show $(X×Y)-G_f$ is open, so let $x×y∉G_f$. Then $f(x)≠y$, so since $Y$ is Hausdorff choose a neighborhood $V$ of $y$ not containing $f(x)$. Now $Y-V$ is closed in $Y$, so since compact Hausdorff spaces such as $Y$ are regular, choose a neighborhood $W$ of $y$ so that in particular $\overline{W}⊆V$. Now, if $x∈\overline{f^{-1}(W)}$, then since $f$ is continuous we derive $$f(x)∈f(\overline{f^{-1}(W)})⊆\overline{W}⊆V$$ so $f(x)∈V$ despite the choice of $V$. Hence choose some neighborhood $U$ of $x$ such that $U∩f^{-1}(w)=ø$. We claim $U×W$ is a neighborhood of $x×y$ disjoint from $G_f$; assume otherwise that $z×f(z)∈U×W$ for some $z∈X$. Then $z∈U∩f^{-1}(W)$, a contradiction.

($⇐$) Suppose $G_f$ is closed. Then let $K⊆Y$ be closed. We see $$f^{-1}(K)=π_1(G_f∩[X×K])$$ since $f(x)∈K$ implies $x×f(x)∈G_f∩[X×K]$, and $x×k∈G_f$ for $k∈K$ implies $f(x)=k∈K$. Now since $Y$ is compact we see projection $π_1$ is a closed map, so $f^{-1}(K)⊆X$ is closed.$~\square$