# Question about a function on a compact metric space

This is the question, I have solved (a) and know that (c) follows immediately from (b). But I cannot work out (b). The furthest I can go is that I know each $f(F_n)$ is compact by continuity of $f.$

Let $X$ be a compact metric space, let $Y$ be a metric space, and let $f : X → Y$ be continuous. Let $\{F_n\}$ be a sequence of nonempty closed subsets of $X,$ which therefore are also compact sets. Suppose that $F_n ⊃ F_{n+1}$ for all $n.$ It’s obvious that $\bigcap\limits_n F_n$ is nonempty.

(a) Prove that $\bigcap\limits_n f(F_n)$ is nonempty.

(b) Let $y ∈ \bigcap\limits_n f(F_n)$. Prove that there is a subsequence $\{F_{n_j} \}$ of $\{F_n\}$ and a corresponding sequence of points $\{x_{n_j} \}$ such that $x_{n_j} ∈ F_{n_j} ,~ f(x_{n_j} ) = y,$ and $\{x_{n_j} \}$ converges to a point $x.$ Prove that $x ∈ \bigcap\limits_n F_n.$

(c) Prove that $\bigcap\limits_n f(F_n) ⊂ f(\bigcap\limits_n F_n).$

You can just choose a sequence $\{x_n\}_{n\in\Bbb{N}}$ such that $x_n\in F_n$ and $f(x_n)=y$ by the assumption $y\in \bigcap\limits_n f(F_n).$ Then all $x_n$ falls in $F_1$ and by the compactness there exists a convergent subsequence $\{x_{n_j}\}.$ The limit falls in the intersection is then clear by the convergence.
In fact, for $n_k\ge n_j,$ we have $x_{n_k}\in F_{n_j},$ which is a closed set since it's a compact set in a Hausdorff space. Thus $x\in F_{n_j}$ for any $j\in\Bbb{N}$ so $x$ is in the intersection.