Let $X$ be a complete thin metric space and let $A, B$ be disjoint closed connected subset of $X$ then there is a compact set $K$ such that each neighborhood of $K$ disjoint from $A \cup B$ separates $A$ and $B$.
Can someone explain how I can prove this? I have already read the proofs that used a sequence but I don't really understand how the sequence suddenly becomes Cauchy.
and here is definition of thin space
A uniform space $X$ is said to be thin if for every closed subset $Y$ of $X$ and every $y \in Y$, the uniform quasi-component of $y$ in $Y$ is connected.
here is the prove that i am trying to understand:
To prove this, let $H_{n+1}$ $\subseteq$ $H_n$, $H_n$ seperates $A$ and $B$, and $H_n$ contained in a finite union of balls of diameter < $1/n$. And $K$ = $\cap_n H_n$
It suffice to prove that every open neighbourhood $U$ of $K$ contains some $H_n$ (why?). We reason by contradiction. So assume that for each n there is $x_n$ $\in$ $H_n -U$. We can then easily extract from $\{x_n\}$ a subsequence $\{y_i\}$ such that for each $n$ all but finitely many of the points $y_i$ lie in only one of the finitely many open balls of the fixed cover of $H_n$. It follows then that $\{y_i\}$ is a Cauchy sequence, hence it converges to a point $y$ which must lie in $K$ and also outside of $U$, a contradiction.