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: 


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*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 $X$ be a compact metric space and let $\{p_n\}$ be a Cauchy sequence in $X$. Then define $E_N$ as $\{p_N, p_{N+1}, p_{N+2}, \ldots\}$. Let $\overline{E_N}$ be the closure of $E_N$. Since it is a closed subset of compact metric space, it is compact as well. 
By definition of Cauchy sequence, we have $\lim_{N\to\infty} \text{diam } E_N = \lim_{N\to\infty} \text{diam } \overline{E_N} = 0$. Let $E = \cap_{n=1}^\infty \overline{E_n}$. Because $E_N \supset E_{N+1}$ and $\overline{E_N} \supset \overline{E_{N+1}}$ for all $N$, we have that $E$ is not empty. $E$ cannot have more than $1$ point because otherwise, $\lim_{N\to\infty} \text{diam } \overline{E_N} > 0$, which is a contradiction. Therefore $E$ contains exactly one point $p \in \overline{E_N}$ for all $N$. Therefore $p \in X$.
For all $\epsilon > 0$, there exists an $N$ such that $\text{diam } \overline{E_n} < \epsilon$ for all $n > N$. Thus, $d(p,q) < \epsilon$ for all $q \in \overline{E_n}$. Since $E_n \subset \overline{E_n}$, we have $d(p,q) < \epsilon$ for all $q \in E_n$. Therefore, $\{p_n\}$ converges to $p \in X$. 
