In order to eliminate degenerate configurations, I suggest adding the constraint that no more than three points lie in the same plane. (This also implies all five points are distinct.) This would better model the case of actually drawing points on a real ball, since there will always be physical imperfections.
With that additional constraint, the claim is
true.
To see this, first:
pick two points that are not antipodes (exactly opposite each other) and whose great circle (which is uniquely determined by two points) does not contain a third point.
This is always possible (or else the problem immediately solved) because:
there can be at most one antipodal pair, because if there were two, then a great circle would contain all four, meaning they also lie on the same plane. If there is one antipodal pair, then if the other three all lie on the same great circle, then the original problem is immediately solved: that great circle plus either of its hemispheres contains four points. If there are no antipodes, then there can be at most two great circles containing three points (imagine five points arranged like an X), occupying 6 out of the 10 pairs and leaving 4 pairs that each determine a great circle that is not shared with any other point.
Now,
having selected two points whose great circle contains no other points, the remaining three points are either divided into two on one half and one on the other, or all three are on the same side. In the former case we select the hemisphere with two, which when combined with the two on the great circle yields four. In the latter case (all three on the same side), twist the circle by an infinitesimal amount around one of the points it contains so the other point previously on the circle is now on the opposite side from the three. Then, again, we have exactly four to a side.
The reasoning in my answer is somewhat similar to that in the answer by AxiomaticSystem (although developed independently), especially the last step. I think the approach based on great circles offers some additional clarity because the sphere is evenly divided into hemispheres at all stages, but this may just be a case of inventor's bias.