If ${E_n}$ is a sequence of closed nonempty and bounded sets in a complete metric space $X,$ if $E_{n+1} \subset E_n,$ and if $\lim_{n\rightarrow \infty}\text{diam}(E_n) =0,$ then $\bigcap_1^\infty E_n$ contains exactly one point.
I am not sure why I need that the sets are bounded.
My proof is as follows: Begin by constructing a sequence $(p_n)$ such that the $n^{th}$ term is an element of $E_n.$ $\lim_{n\rightarrow \infty}\text{diam}(E_n)=0,$ for each $\epsilon>0$ there is an $N$ such that if $n\geq N$, $|\text{diam}(E_n)-0| <\epsilon.$ Fix N. This implies if $p$ and $q$ are in $E_N$, $d(p,q)<\epsilon$. By the way we constructed our sequence, we know that there is some $p_N\in E_N$ and for all $p_{N'}$ and $p_{M'}$ where $N'\geq N',M'\geq N$, $d(p_{N'},p_{M'})<\epsilon$ because $p_{N'}\in E_{N'} \subset E_{N}$ and $p_{M'}\in E_{M'} \subset E_{N}$. Thus this sequence is Cauchy.
We are given that $X$ is complete, thus every Cauchy sequence converges. Let $\lim_{n\rightarrow\infty}p_n=p$. We know every open ball around $p$ must contain all but finitely many elements of the sequence--thus it must contain some element of each $E_n$. Therefore $p$ is a limit point of each $E_n$ since the sets are nested. Since each $E_n$ is closed, $p$ must be an element of each $E_n$. Thus $\bigcap_1^\infty E_n$ is nonempty.
At this point we just need to show that $p$ is the only element in the grand intersection. Assume via contradiction that there exists more than one point $p$ in the grand intersection. Call it $q$. Let $L=d(p,q)$. Choose $\epsilon
I didn't use that the sets were bounded anywhere, which leads me to believe this logic is in some way flawed.