I am trying to prove:
If $\{E_n\}$ is a sequence of closed, nonempty, and bounded sets in a complete metric space $X$, if $E_n \supset E_{n+1}$, and if $$\lim\limits_{n \to \infty} \textrm{ diam } E_n = 0,$$ then $\cap_{1}^{\infty} E_n$ consists of exactly one point.
Here is my attempt:
Proof: Suppose $\{E_n\}$ is a sequence of closed, nonempty, and bounded sets in a complete metric space $X$ such that $E_n \supset E_{n+1}$ and \begin{equation}\label{21.1} \lim\limits_{n \to \infty} \textrm{ diam } E_n = 0 \end{equation} Denote $E = \cap_{1}^{\infty} E_n$. Suppose to the contrary that $E$ does not contain exactly one point. Then, either $E$ contains more that one point or is empty. If $E$ contains more that one point, then $\textrm{ diam } E > 0$. But, for each $n$, $E_n \supset E$, so that $\textrm{ diam } E_n \geq \textrm{ diam } E > 0$ which contradicts $\lim\limits_{n \to \infty} \textrm{ diam } E_n = 0$. This shows that $\left|E\right|\leq 1$.
If $E$ is empty, then each non-empty $E_1, E_2, \dots$ must be mutually disjoint. This contradicts $E_n \supset E_{n+1}$ and we are done.
My question: I never used the fact that $X$ is a complete metric space and that each $E_n$ is nonempty and bounded. This makes me think that there is something wrong with my proof. Can someone please let me know what's wrong with my proof?
Now, construct a sequence $\{p_n\}$ such that $$p_1, \in E_1, p_2, \in E_2, \dots .$$ We know that each $p_n$ exists since each $E_n$ is non-empty. Further, we claim that $\{p_n\}$ is Cauchy in $X$.
[Some argument that involves $\lim\limits_{n \to \infty} \textrm{ diam } E_n = 0$, perhaps, which proves that $\{p_n\}$ is indeed Cauchy in $X$.]
Since $X$ is complete, $\exists p \in X$ such that $\{p_n\} \to p$. Since each $E_n$ is complete, $p \in E_n$ for each $n$, that is, $p \in E$. This shows that $\left|E\right|\geq 1 \implies \left|E\right| = 1$.
My question: How can I prove that $\{p_n\}$ is Cauchy? Thanks!
Based on @Daniel Fisher's and @TSU's comments:
Proving that $\{p_n\}$ is Cauchy in $X$: Notice that $p_n, p_{n+1}, \dots$ all belong to the set $E_n$. Without loss of generality, assume that $p_b, p_c \in E_n$ are such that $b , c \geq n$ and $\textrm{diam } E_n = d(p_b, p_c)$. Now, recall that $\lim\limits_{n \to \infty} \textrm{ diam } E_n = 0$ which means that for all $\epsilon > 0$ \begin{equation*} n \geq N \implies d(p_b, p_c) < \epsilon \end{equation*} Here, we have shown that for all $\epsilon > 0$, $\exists N \in \mathbb{N}$ such that $b , c \geq n \geq N \implies d(p_b, p_c) < \epsilon$ by the definition of the diameter of $E_n$. This shows that $\{p_n\}$ is Cauchy in $X$.