Suppose the graph is closed and $V$ is open in $Y$ . If $V\cap f(X)=\emptyset$ then $f^{-1}(V)$ is empty and open. Now suppose that $V$ is a neighborhood of $f(x)$ for some $x$ . Then $X\times V$ is open in the product and $X\times(Y-V)$ is closed and so is its intersection with the graph. Using Exercise 7, its image under the projection on $X$ is closed as well, but the complement of the image is the preimage $f^{-1}(V)$ , and it is open. Therefore, $f$ is continuous. Note that we use compactness here, as Exercise 7 requires it.

Now, suppose that $f$ is continuous. We take a point $(x,y)$ outside of the graph and show that there is its open neighborhood that does not intersect the graph. This will show that the complement of the graph is open, and, hence, the graph is closed. For points $f(x)$ and $y\neq f(x)$ take two disjoint open neighborhoods $V$ and $V'$ , respectively, using Hausdorff property of $Y$ . Then, since $f$ is continuous, $U=f^{-1}(V)$ is open, $x\in U$ , and $U\times V'$ is an open neighborhood of $(x,y)$ such that it does not intersect the graph (as $(a,f(a))\in U\times V'$ implies $f(a)\in V\cap V'$ ). Note that this direction does not use compactness of $Y$ .