Let $X$ and $Y$ be topological spaces where $Y$ is compact and Hausdorff. Then let $f:X\rightarrow Y$ be a function and $G_{f}=\{(x,y)\in X\times Y|y=f(x)\}$ be the graph. The theorem is stated as such:
"$f$ is continuous $\Leftrightarrow$ $G_{f}$ is closed in $X \times Y$"
Here I am only interested in the $\Leftarrow$ direction. This particular incarnation of this problem comes with a hint, but I am having a hard time moving forward with it. This is what I have so far;
"Let $x_{0}\in X$ where $V=V(f(x_{0}))$ is a neighborhood of $y=f(x_{0})$. Then there exists a neighborhood of $x_{0}$, say $U$, such that $f(x_{0})\in V$."
My main difficulties arise in the next statement
"Does there exist a tube $U \times Y$ with the slice $\{x_{0}\} \times (Y\setminus V)$ which does not intersect $G_{f}$?"
I cannot see how this ties into the continuity of $f$. I see that we begin with any fixed point in $x$ take the neighborhood around it and build a tube in the product space (since $Y$ is compact) then we remove this point's image and neighborhood from the tube and ask if it (the tube I assume) does not intersect the respective graph or rather does such a tube exist.
I am having a hard time answering the above statement and seeing how that answer ties into the continuity of $f$. Some guidance from here would be very helpful.
Try to show that the projection $\pi: X \times Y \to X$ is a closed map (from compactness of $Y$ and the "tube lemma") and use that to show continuity of $f$ via inverse images of closed sets being closed.