Nested sequences of balls in a Banach space This seems to be a fairly easy question but I'm looking for new points of view on it and was wondering if anyone might be able to help.
(By the way- this question does come from home-work, but I've already solved and handed it, and I'm posting this out of interest, so no HW tag.)

Let $B_n=B(x_n,r_n)$ be a sequence of nested closed balls in a Banach space $X$. Prove that $\bigcap_{n=1}^\infty B_n\neq\varnothing$.

As I said before, it should be rather simple. When the radii decrease to 0, it's just a matter of selecting any sequence of points in $B_n$, and it must be Cauchy- and the limit is in the intersection.
My question is what to do when the radii do not decrease to 0? I got some tips about multiplying the balls by a sequence of decreasing scalars, or reducing the radii so that they decrease to 0, but found too many pathological cases for both methods.
Finally- I used a geometric arguemnt (which i've shown to work in any normed space) that if $B(x_1,r_1)\subset B(x_2,r_2)$ then $\| x_1-x_2\|\leq|r_1-r_2|$.
This turned out to be some kind of technical catastrophe, but it worked...
Still, if anyone knows of a more elegant solution, I'd love to hear about it.
 A: I don't know if this is more elegant, but that's about the best I can come up with at the moment and probably essentially the same as your argument.

Consider first the situation $B_{\leq r}(x) \subset B_{\leq s}(y)$. It is easy to see that $r \leq s$.
Claim. $\|y - x\| \leq s - r$.
Proof. If $x = y$ there is nothing to prove, so let's assume $x \neq y$. The point $z = x - r \frac{y-x}{\|y - x\|}$ belongs to $B_{\leq r}(x)$ and hence also to $B_{\leq s}(y)$. Therefore $\|y - z\| \leq s$. On the other hand,
\[
y - z = y - x + \frac{r}{\|y - x\|} (y - x) = \underbrace{\left(1 + \frac{r}{\|y - x\|}\right)}_{\lambda} (y - x),
\]
so $s \geq \lambda \|y - x\| = \|y - x\| + r$ and hence $\|y - x\| \leq s - r$. 

This means that a nested sequence of closed balls $B_{\leq r_{n}}(x_{n})$ has the following properties:


*

*The sequence $r_{n}$ is monotonically decreasing, hence converges to some $r$.

*If $N$ is such that $r_{N} \leq r + \varepsilon$ then the above claim implies that for all $n\geq m \geq N$ we have $r_m - r_n \leq \varepsilon$, so $\|x_{m} - x_{n}\| \leq \varepsilon$ because $B_{\leq r_{n}}(x_{n}) \subset B_{\leq r_{m}}(x_{m})$.


In other words, the centers $x_{n}$ form a Cauchy sequence and their limit point $x$ must belong to $\bigcap_{n = 1}^{\infty} B_{\leq r_{n}}(x_{n})$.

Added: As Jonas pointed out, the argument can be made even simpler and doesn't need completeness: Suppose $r_{n} \to r \gt 0$. Then there is $N$ such that $r_{N} \leq 2r$. Then for all $n \geq N$ we have $r \leq r_{n} \leq r_{N} \leq 2r$, so $r_{N} - r_{n} \leq r$ and the claim implies that $\|x_{n} - x_{N}\| \leq r \leq r_{n}$, so $x_{N} \in \bigcap_{n = 1}^{\infty} B_{\leq r_{n}} (x_{n})$.
