If $\{E_n\}_{n \in N}$ is a sequence of closed nonempty and bounded sets in a complete metric space $(X,d)$, if $E_n \supset E_{n+1}$ and if $$\lim_{n \to \infty} \operatorname{diam} E_n = 0,$$ then $$\bigcap_{n=1}^{\infty} E_n$$ consists of exactly one point.
Attempt:
Since $$\lim_{n \to \infty} \operatorname{diam} E_n = 0,$$ this implies $$\lim_{n \to \infty} \sup_{n \in \mathbb{N}} (d(p_n,q_n)) = 0$$ for $p_n,q_n \in E_n$. Therefore, $$\lim_{n \to \infty} d(p_n,q_n) = 0$$ so that $$\lim_{n \to \infty} p_n = x = \lim_{n \to \infty} q_n.$$ by definition of a metric. Thus, since $E_n \supset E_{n +1}$, we see that $x \in E_n$, for all $n \in \mathbb{N}$. Thus, $$\bigcap_{n=1}^{\infty} E_n = \{x\}$$ as desired.
I believe this proof is incomplete, since I didn't use the fact that the $E_n$'s are closed, or that $X$ is complete. Where am I mistaken? Please, no complete answers, as this is homework.