Exercise 1.21 of An Introduction to Banach Space Theory by Megginson:
Let $X$ be normed space. If $(B_n)$ is a nested sequence of closed balls in $X$, then the centers of the balls form a Cauchy sequence.
How to prove it?
It also asked for a counterexample for metric space, which is easy: in an ultrametric space, like the $p$-adic number field, any point in a ball is a center of the ball.
If, in a normed space, you have two balls, $B_r(a)$ and $B_{r'}(a')$ and if $B_r(a)\supset B_{r'}(a')$, then $\|a-a'\|$ is $r-r'$, at most. You can prove it as follows (see the picture below): if $a=a'$, it's trivial. Otherwise, consider the ray whose origin is $a$ and passes through $a'$. Let $p$ be the point of the ray whose distance to $a$ is $r$ (that is, $p=a+\frac r{\|a-a'\|}(a-a')$). Then $a$, $a'$ and $p$ are colinear, with $a'$ being between the other two. So$$\|p-a\|=\|p-a'\|+\|a'-a\|.$$Therefore $r=\|p-a'\|+\|a-a'|\geqslant r'+\|a-a'\|$ and so $\|a-a'\|\leqslant r-r'$.
Now, if $r_n$ is the radius of $B_n$, then $(r_n)_{n\in\mathbb N}$ is a decreasing sequence of non-negative real numbers. Therefore, it converges. So, given $\varepsilon>0$, $|r_m-r_n|<\varepsilon$ if $m,n\gg1$. But then the distance between the centers of $B_m$ and $B_n$ is smaller than $\varepsilon$.
Let $\{\bar{B}(x_{n},r_{n})\}$ be a decreasing sequence of closed balls. We have $x_{n+1}+\frac{r_{n+1}}{\left\Vert x_{n+1}-x_{n}\right\Vert }% (x_{n+1}-x_{n})\in \bar{B}(x_{n+1},r_{n+1})\subseteq \bar{B}(x_{n},r_{n})$ so $\left\Vert x_{n+1}+\frac{r_{n+1}}{\left\Vert x_{n+1}-x_{n}\right\Vert }% (x_{n+1}-x_{n})-x_{n}\right\Vert \leq r_{n}$ which says $\left\Vert x_{n+1}-x_{n}\right\Vert +r_{n+1}\leq r_{n}$ i.e. $\left\Vert x_{n+1}-x_{n}\right\Vert \leq r_{n}-r_{n+1}$. This implies that $\{r_{n}\}$ is decreasing ( which is also obvious from the fact that the diameters of $% \bar{B}(x_{n},r_{n})$ are decreasing). Let $r_{n}\downarrow r$. Iteration of $\left\Vert x_{n+1}-x_{n}\right\Vert \leq r_{n}-r_{n+1}$ yields $\left\Vert x_{n+m}-x_{n}\right\Vert \leq r_{n}-r_{n+m}\rightarrow 0$ as $n,m\rightarrow \infty $
