 Solution

The inclusion $f({\bigcap }_n{K}_n)\subseteq {\bigcap }_{n}f(K_n)$ holds for every map. Conversely, let $y$ belong to the right side. The sets $L_n=K_n\cap f^{-1}(\{y\})$ are nonempty and nested. They are compact because singletons are closed in a Hausdorff space and $f$ is continuous. By part (b)(i), some $x$ lies in every $L_n$. Then $x\in {\bigcap }_n{K}_n$ and $f(x)=y$, proving equality of the two sets.