Every compact metric space is complete I need to prove that every compact metric space is complete. I think I need to use the following two facts: 


*

*A set $K$ is compact if and only if every collection $\mathcal{F}$ of closed subsets with finite intersection property has $\bigcap\{F:F\in\mathcal{F}\}\neq\emptyset$.

*A metric space $(X,d)$ is complete if and only if for any sequence $\{F_n\}$ of non-empty closed sets with $F_1\supset F_2\supset\cdots$ and $\text{diam}~F_n\rightarrow0$, $\bigcap_{n=1}^{\infty}F_n$ contains a single point.


I do not know how to arrive at my result that every compact metric space is complete. Any help? 
Thanks in advance.

Let $\langle F_n\rangle_{n\in\Bbb{N}}$ be a descending sequence of nonempty closed sets satisfy that $\operatorname{diam} F_n\to 0$ as $n\to\infty$. You can easily check that if $m_1<m_2<\cdots<m_k$ then 
$$
F_{m_1}\cap F_{m_2}\cap\cdots\cap F_{m_k} =F_{m_k}\neq \varnothing
$$
so $\langle F_n\rangle_{n\in\Bbb{N}}$ satisfies finite intersection property. Since $(X,d)$ is a compact metric space, $\bigcap_{n\in\Bbb{N}} F_n$ is not empty. Since 
$$
\operatorname{diam} \bigcap_{n\in\Bbb{N}} F_n \le \operatorname{diam} F_m\to 0\qquad \text{as }\> m\to\infty
$$
so $\bigcap_{n\in\Bbb{N}} F_n$ contains at most one point. So $\bigcap_{n\in\Bbb{N}} F_n$ is singleton.
