Let $(X,d)$ be a metric space. Let $\mathcal{C}$ be the set of all collections $\{O_i\}_{i=1}^\infty$ of non-empty closed subsets such that \begin{align*}
&(a) O_{n+1}\subset O_n \forall n \\
&(b) \lim\operatorname{diam} (O_n) = 0 \ as \ n \to \infty
\end{align*}
Prove that $X$ is complete if and only if $\forall C \in \mathcal{C}$ \begin{align*}
\bigcap_{A\in C} A \not= \emptyset
\end{align*}
For the $if$ part: For every $n$, choose $x_n \in O_n$. Then since $O_{n+1}\subset O_n$, the set $\{x_n,x_{n+1},x_{n+2},\cdots\}\subset O_n$. Since $\lim\operatorname{diam}(O_n) = 0$, for any $\epsilon > 0$ choose a natural number $N$ so that $\operatorname{diam}(O_n)<\epsilon$ for $n\geq N$. This means that for any $n,m \geq N$, $|x_n-x_m| \leq\operatorname{diam}(\{x_{N},x_{N+1},\dots\}) \leq\operatorname{diam}(O_N) < \epsilon$. So $\{x_n\}$ is a Cauchy sequence. By completeness of $X$, $\{x_n\}$ converges to a point, $a$. By $O_n$ being closed it must contain $\{x_n,x_{n+1},\dots\}$. Thus, for any given $D\in\mathcal{C}$ we get
\begin{align*}
a\in \bigcap_{A \in D} A.
\end{align*}
hence non-empty.
Also, I am not sure how to start the other direction. I think you need to look at the tails of each sequence
Best Answer
Consider the Cauchy sequence $\{x_n\}$. Define $$F_n :=\overline{ \{x_m \vert m \geq n\}}$$.Note that $F_n $ is closed and satisfies the given two conditions. Let $\mathcal{F} = \{F_n\}_{n=1}^\infty$. Then $$F=\bigcap_{F_n\in \mathcal {F} } F_n \not= \emptyset$$. Take $x\in F $ and prove that $x $ indeed is the limit of the sequence $\{x_n\}$. You may have to use the fact that $d(A)= d(\overline {A})$, where $d(A) $ is the diameter of the set $A $ and $\overline {A}$ is the closure of $A $.