Let $S$ be the finite set of these points, situated on the surface of a sphere of centre $O$. Consider their convex hull $\operatorname{conv}(S)$. It is easy to see that $O\not \in \operatorname{conv}(S)$ if and only if $S$ is contained on the surface of an open hemisphere (by well-known results on separability of a point from a convex set). Assume $O\in \operatorname{conv}(S)$. By the Caratheodory theorem, there exists a subset $T$ of $4$ points in $S$, such that $O\in \operatorname{conv}(T)$. By the given condition we however have that $T$ is contained on the surface of an open hemisphere, thus $O\not \in \operatorname{conv}(T)$, contradiction. Therefore $O\not \in \operatorname{conv}(S)$, and so by the previous remark $S$ is contained on the surface of an open hemisphere. Notice the result stands no more for any infinite set of points, even a countable one. Say one point in $S$ lies on the equator, and each other point of the equator does not lie in $S$, but is an accumulation point for $S$. Then $S$ is not contained on the surface of an open hemisphere, but any $4$ points of it are.