Let $X$ be the unit sphere in $\ell^2$, i.e. $X=\{x\in\ell^2: \|x\|=1\}$. Let the metric on $X$ be the geodesic metric, i.e. $d(x,y)=\cos^{-1}\langle x,y\rangle$. Call a set a ball-intersection if it is an intersection of closed balls with centers in $X$. 


Does there exist a decreasing sequence of nonempty ball-intersections in $X$ with void intersection?


If we let $S\_0=\cap\_i B[e\_i,\pi/2]=\{x\in X: x\_i\ge 0 \forall i \},$ $S\_1=\{x\in S\_0: x\_1=0\},$ $S\_2=\{x\in S\_0: x\_1=x\_2=0\}, \cdots$, then $\cap\_i S\_i=\emptyset$. But $S\_i$ for $ i\neq 0$ are not ball-intersections.



Geometry of the Hilbert sphere
I think yes. Your balls are of the form $X\cap S(x,r)$ where $x\in X$ and $S(s,r)$ is the slice
$\{y: \|y\|\le 1 \ \text{and} \ \langle x,y\rangle \ge r\}$ of the unit ball. Note that if $y$ is in a slice, so is $y/\|y\|$. The slices have non empty intersection because they are weakly compact.


EDIT: This argument looks OK if all $r$ are positive, but $r$ can be negative, so some further thought is needed.


