 Solution

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.