I know this question has asked to death here on MSE but I have not found a satisfactory solution. A solution found online is extremely elegant but I do not quite understand it!
Given nested sequence of closed nonempty connected subsets of a compact metric space $X$. Prove that $\bigcap_{i=1} X_i$ is nonempty and connected.
It is a standard fact that arbitrary intersection of compact set is nonempty and compact.
I wish to understand the bit of the proof that this is also connected
Proof: https://math.berkeley.edu/sites/default/files/pages/f10solutions.pdf
Suppose that $\bigcap_{i=1} X_i$ is not connected. Let $A$ and $B$ be two disjoint nonempty closed sets so that $\bigcap_{i=1} X_i = A \cup B$. Find disjoint open sets $U$ and $V$ so that $A \subset U$ and $B \subset V$.
Put $F_i = X_i − (U \cup V )$. Then $\{Fi\}$ is a nested sequence of compact sets, whose intersection is empty. Thus $F_i = ∅$ for some $i$. That is, $X_i \subset U \cup V$.
However, $X_i$ intersects both $U$ and $V$ , since $X_i \cap A\neq ∅$ and $X_i \cap B \neq ∅$. This contradicts the assumption that $X_i$ is connected.
Can someone please elaborate on some of the details of the proof?
1) Why bother finding open sets containing $A,B$ and how do we know that they even exist? This move seems sort of unnatural.
2) What is so special about $U \cup V$ especially, why does $F_i = X_i − (U \cup V )$ imply $\{F_i\}$ has empty intersection? Why does intersection being empty imply the existence of $F_i = \varnothing$?
3) Can someone please justify $F_i = \varnothing \Leftrightarrow X_i \subset U \cup V$.
4) Can someone please justify the contradiction?
I know this is the pretty much the entire proof, but I have seriously tried to crack it and failed to understand it. Perhaps the proof is too difficult.
