5 points on a ball, divide the ball into 2 halves so that one half as exactly 4 points

Kudos to loopy for getting the answer. Some reasoning to show that it's essentially best possible:
Take a sphere and cut it generally (i.e. through no points). If four points lie on a side, we are done.
If five lie on one side, then choose a polar axis through the cut such that there is a unique point with least longitude: there are finitely many cases where this is not possible - one for each side of the convex hull of the five points - so this is possible to do. Once an axis is found, simply rotate the cut plane around it until it passes the nearest point and the four others are isolated, as desired.
Assume now that three lie on one side and two on the other. Make a new cut plane through the two isolated points. If this plane divides the three such that two lie on one side, we are done. If three all lie on one side and none on the other, then we can twist the cut plane slightly such that one of the two moves to the other half, and we are done. The only remaining case is if the three all lie on the plane defined by the other two, in which case division of the sphere is only possible if a half of the circle containing all five points contains exactly four points.

The claim is

True.

If you draw 4 points, you can grab any 3 of them and grab the hemisphere that contains that 3 points. So, if you draw 1 point, then you can grab 3 points so that the hemisphere that contains that 3 points can contain the last 1 point.

The claim is

Edit:

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

To see this, first:

This is always possible (or else the problem immediately solved) because:

Even more detail on the previous point (skip this at first):

Now,

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.

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Finally, let's consider the case between my suggested constraint (no more than three in a plane) and that of loopy's answer (all five in a plane): exactly four line in a plane.

In this case the claim is:

Case 1:

Case 2: