Minimum number of hemispheres covering a sphere

Question

Here is a question which seems easy but seems to have many pitfalls. If I give you an arbitrary covering of the sphere by $N$ closed hemispheres. You can pick any of the hemispheres to keep. What is the minimum number you can keep while still covering the sphere? We suspect the answer is $4$ but we can't seem to prove it.

Answer 1

Hint: Look up Helly's theorem.

EDIT: I should have been more specific, since Helly had more than one theorem. The one I'm talking about is the one found at http://en.wikipedia.org/wiki/Helly%27s_theorem If your closed hemispheres are $H_1, \ldots, H_n$ (in the sphere $S$), let $X_i$ be the convex hull of $S \backslash H_i$. If no $4$ of your hemispheres cover the whole sphere, that says every $4$ of the $X_i$ have nonempty intersection, and then (since we're in ${\mathbb R}^3$) Helly says the intersection of all the $X_i$ is nonempty, and that implies that $H_i$ don't cover the whole sphere.