 Solution

The product topology has basis $U\times V$ with $U$ open in $X$ and $V$ open in $Y$. Since ${\Pi }_X^{-1}(U)=U\times Y$ and similarly for ${\Pi }_Y$, both projections are continuous.

If $Y$ is Hausdorff, a point $(y_1,y_2)$ off the diagonal has disjoint product neighborhoods, so the diagonal is closed. Conversely, if the diagonal is closed, its complement supplies disjoint neighborhoods of any two distinct points.

The graph is the inverse image of the diagonal under $(x,y)\mapsto (f(x),y)$, so a continuous map into a Hausdorff space has closed graph. Conversely, if $X,Y$ are compact Hausdorff and $C\subseteq Y$ is closed, then ${\Gamma }_f\cap (X\times C)$ is compact. Its projection onto $X$ is $f^{-1}(C)$, compact and hence closed. Thus $f$ is continuous.

The function $f(0)=0$, $f(x)=1/x$ for $x\ne 0$ is discontinuous but has closed graph: a convergent graph sequence approaching $x=0$ cannot have $x\ne 0$ and bounded second coordinate unless it eventually reaches the isolated graph point $(0,0)$.