Suppose $\left(V,\left\|\cdot\right\|\right)$ is a strictly convex normed space, $S\subseteq V$ and $x\in V$. Then, iff $x$ is in the closed convex hull of $S$, we have $$\bigcap_{a\in S}B\!\left(a,\left\|x-a\right\|\right)=\varnothing,$$ where $B$ denotes open balls. Is this true? How to prove or construct a counterexample?
I can easily prove this when $S$ consists of two distinct points. It also seems correct when $S$ are discrete points in the Euclidean plane. I wonder whether it can be generalized to more general $V$ and $S$.
For the case when $x$ is not in the closed convex hull of S, here is my idea to prove that there is some point $y$ in the intersection of the open balls. There exists a closed half-space $Y$ of $V$ such that $S$ and $x$ are separated by it and that $S$ does not contain a sequence that approaches any point in $Y$. By Riesz's lemma, we can choose a point $y$ in $Y$ whose distance to $x$ is arbitrarily close to the distance $d$ from $x$ to $Y$. My idea is to prove that, when $\left\|y-x\right\|$ is small enough (close enough to $d$), we have $\left\|y - a\right\|<\left\|x - a\right\|$ for any $a\in S$, then $y$ is a point in the intersection of the open balls. However I do not know how to proceed.
For the case when $x$ is in the closed convex hull of $S$, I currently do not have any idea.