First take a sphere S2 in 3D space. Next consider the free group G generated by 2 elements, i.e. <a,b> which consists of all possible combinations of a, b, a-1, b-1. Each element of G can be written in a unique way. G can be split into 5 disjoint sets: G = {e} U S(a) U S(a-1) U S(b) U S(b-1) where S(x) represents the set of elements in G whose 'string' starts with x.
It is clear that aS(a-1) is equal to {e} U S(a-1) U S(b) U S(b-1) (the a 'cancels' the a-1 at the start of each element in S(a-1) leaving all elements starting with either a-1, b or b-1. Similarly bS(b-1) is equal to {e} U S(b-1) U S(a) U S(a-1).
So aS(a-1) U S(a) = G. Same for bS(b-1) U S(b).
Now consider a and b to be rotations of the sphere about orthogonal axes by some irrational number of degrees - say 1 radian. Every element of G is a combination of rotations about these 2 axes, and is thus a rotation. Each element of G (except e) will therefore have 2 fixed points on the sphere. Let the set of all these fixed points be Q, which is countable (as the elements of G are countable, being...equinumerous with N?...Wait! This is wrong!
I thought I understood this, but it doesn't work. The elements of G aren't countable, are they? Perhaps they are. Back to the drawing board. Rest of the proof involves the action of G on the set of points in the sphere, writing the set of points as a disjoint set of the orbits of G acting on S-Q, invoking the axiom of choice to select one point from each orbit which are put into a set M. This set is then composed with the disjoint bits of G to decompose the point-set of the sphere into 4 bits, two of which are translated across (say MS(a) and MS(a-1)) and then MS(a-1) is rotated by rotation a, and hey presto you've got your original sphere back. Do the same with the non-translated bits and you've got 2 spheres.
Few technical details with regard to the set Q - you have to rotate it by something not a member of G or something like that, and to extend the transformation to a solid ball rather than a hollow sphere you have to include the centre as a separate point and use the projections of each point to the centre, but it's largely the same.
It's all nonsense anyway - axiom of choice is clearly false, certainly when it comes to maths that could be applied to the real world. At least that's my opinion.