Nested sequences of balls in a Banach space
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 $x_{N} \in \bigcap_{n = 1}^{\infty} B_{\leq r_{n}} (x_{n})$.