Past exam of the mathematics course of the University of Cambridge / 2009 / ib / Paper 1 / 12F (Metric and Topological Spaces)

Suppose $(x_n,f(x_n))$ converges to $(x,y)$. Continuity gives $f(x_n)\to f(x)$, and uniqueness of limits in a metric space gives $y=f(x)$. Hence every convergent sequence in the graph has its limit in the graph, proving ${\Gamma }_f$ is closed.

If $f$ were not continuous at $x$, there would be $\epsilon >0$ and $x_n\to x$ with $d_Y(f(x_n),f(x))\ge \epsilon$. A compact metric space is sequentially compact, so some subsequence $f(x_{n_k})$ converges to $y$. The closed graph then contains its product limit $(x,y)$, forcing $y=f(x)$, contrary to the lower bound. Thus $f$ is continuous, the compact-codomain form of the closed graph theorem for compact spaces.

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.