Intersection of subsets is a single point 
Theorem:  Let $\{E_n\}$ be a family of closed, non-empty and limited subsets of a complete metric space $(X,d)$. If $E_n\supset E_{n+1}$ and
  $\text{diam}(E_n)\rightarrow 0$ show that  $$\cap_{n\in\mathbb{N}}
 E_n$$ is a single point.

Some definitions are

We say a metric space is complete if every Cauchy sequence converges.
  A sequence is Cauchy if eventually the terms in the sequence get
  arbitrarily close to each other (definition of a complete metric space).

The diameter definition

If $E\subset X$, $E\neq \emptyset$ and $S$ is such that
  $$S=\{d(p,q)\quad;\quad p,q \in E\}$$ then $diam(E)=\sup S$

How I can proof this?
What I think is
If $E_n$ is complete then $d(x_n,x_{n+1})\rightarrow 0$ right? It's already doesn't means that the sequence is a single point?

You pick $x_n \in E_n $. Looking at sequence $\{x_n\},$ it is clear that $x_n ,x_{m} \in E_n$, for all $m > n$ So $d(x_n,x_{m}) \leq diam(E_n) \rightarrow 0.$ Therefore , $\{x_n\}$ is Cauchy sequence in $X$ an so it converges to a point, say $x \in X$. Further $x \in  E$ since for all $m \in N$ the sequence $\{x_n\}$ eventually lies in $E_m$ and $E_m$ is closed.   
Furthermore,  $diam(\cap_{n\in\mathbb{N}} E_n ) \leq diam(E_n) \rightarrow 0.$ So $diam(\cap_{n\in\mathbb{N}} E_n ) = 0$ which implies $\cap_{n\in\mathbb{N}} E_n=\{x\}$
