Sequence of nested balls in Metric space $X$ is complete iff... I need to show that a metric space $X$ is complete iff for every nested sequence of closed balls in $X$ with radii tending to zero has nonempty intersection.
A sequence $(B_n)$ of balls is said to be nested if $ B_{n}\supset B_{n+1}$ for all $n$.
Please help me get started.
Thank you.

$\newcommand{\cl}{\operatorname{cl}}$HINT: The easier direction is to show that if $X$ is complete, every nested decreasing sequence of closed balls has non-empty intersection. Let $\langle B_n:n\in\Bbb N\rangle$ be such a sequence, where $B_n$ is the closed ball of radius $r_n$ and centre $x_n$. You know that $\lim_{n\to\infty}r_n=0$; use this to prove that $\langle x_n:n\in\Bbb N\rangle$ is a Cauchy sequence, and that the limit of this sequence is in every $B_n$.
For the other direction, suppose that every nested sequence of closed balls has non-empty intersection, and let $\langle x_n:n\in\Bbb N\rangle$ be a Cauchy sequence. For $n\in\Bbb N$ let $T_n=\{x_k:k\ge n\}$, the $n$-th tail of the sequence. For $n\in\Bbb N$ let $d_n$ be twice the diameter of $T_n$, and let $B_n$ be the closed ball of radius $r_n=2d_n$ centred at $x_n$; then $T_n\subseteq B_n$.
Because the sequence is Cauchy, you know that $\lim_{n\to\infty}d_n=0$. Let $n\in\Bbb N$ be arbitrary. There is a $k>n$ such that $d_k<\frac12d_n$. Suppose that $y\in B_k$; then
$$d(y,x_n)\le d(y,x_k)+d(x_k,x_n)\le r_k+d_n=2d_k+d_n<2d_n=r_n\;,$$
so $y\in B_n$, and it follows that $B_k\subseteq B_n$. Use this observation to construct a nested subsequence of $\langle B_n:n\in\Bbb N\rangle$. By hypothesis this subsequence will have non-empty intersection $A$. Show that $A$ contains only one point, and show that $\langle x_n:n\in\Bbb N\rangle$ converges to this point.
