How do you prove the converse Cantor's intersection theorem? I was given a proof in class on how to prove it in one direction, but I was wondering: is there a way you could prove that if the intersection of a decreasing sequence of closed sets with diameters tending to zero is nonempty, then the metric space is complete?
 A: Take a Cauchy sequence $\langle a_n\rangle $, and use the Cantor intersection property with the sets $\overline {A_n}$ where $$A_n=\{a_n,a_{n+1},\dots\}$$

ADD Firstly, we have to use that for any set $A$, $$\operatorname{diam}A=\operatorname{diam}\bar A$$
For each $\epsilon >0$ there exists $N$ such that whenever $p,n>N$ we have $$|a_{n+p}-a_n|<\epsilon$$
Thus   $\operatorname{diam}A=\operatorname{diam}\bar A\to 0$. By Cantor, $$\exists !x\in\bigcap_{n=1}^\infty \overline{A_n}$$
Take $\epsilon >0$. We want to show there exists $N$ such that $$|a_n-x|<\epsilon$$ whenever $n>N$. 
Take $N$ large enough so that $m>n>N$  gives $$\operatorname{diam}\overline {A_m}<\operatorname{diam}\overline {A_n}<\frac\epsilon 2$$  
Then if $n>N$, $$|x-a_n|\leq |x-a_m|+|a_m-a_n|<\epsilon \;\; \blacktriangle$$
A: Let $\{x_n\}$ be a Cauchy sequence, and for every $n$ let $i(n)$ be the least $n$ guaranteed by the definition of a Cauchy sequence such that $k,m\geq i(n)$ implies $d(x_k,x_m)<\frac1n$.
Now let $C_n$ be the closed ball $\overline B(x_{i(n)},\frac2n)$. Clearly this is a decreasing sequence of closed sets.
If its intersection is non-empty then it is some $x$ such that $\lim x_n=x$. Therefore every Cauchy sequence is convergent.
