Compactness exercises from Dugundji

Solution 1:

You say (under attempt Q1)

Let $y\in \bigcap_{n=1}^{\infty}f[F_n]$ which implies that there exists $x\in\bigcap_{n=1}^{\infty} F_n$ such that $f(x)=y.$

No, this would follow from $y \in f[\bigcap_{n=1}^{\infty} F_n]$ but this is not given, it's what you have to show (still).

Better attempt: for each $n$ we have that $F'_n:=f^{-1}[\{y\}] \cap F_n \neq \emptyset$, as $y \in f[F_n]$ from the intersection. Now apply the FIP property to those smaller compact sets $F'_n$.

For Q2 just apply Zorn's lemma to the poset of "all non-empty compact $A \subseteq X$ such that $f[A]=A$, ordered by reverse inclusion". For a chain in this poset apply the idea from Q1 to see that the intersection of the chain is an upperbound. A maximal element of this poset is as required. Doing the sequence $A_n$ will work too (but do a better job of showing the decreasingness of the $A_n$ eg by induction), but I like Zorn too. And on general principles, it's nice to use Q1 to do Q2. It's a lemma of sorts, why would you want to avoid it?