 Solution

Take $f(0)=0$ and $f(x)=1/x$ for $x\ne 0$. Its graph is $\{(x,y):xy=1\}\cup \{(0,0)\}$, a union of two closed sets, since multiplication is continuous. But $f(1/n)=n$ does not tend to $f(0)$, so the function is not continuous.