So would I. I think it's wrong; I'm pretty sure that the Hausdorff paradox, which is kinda-sorta the equivalent on the sphere, requires some Choice. But maybe I'm confused.

On the other hand, you don't need any Choice for the Sierpinski-Mazuriewicz paradox, which says that there's a subset A of the plane, which can be written as the disjoint union of two "smaller" subsets B and C, but where A, B, and C are all congruent. (But A is countable, so this isn't nearly as startling as Banach-Tarski.) Specifically, let T be translation by (1,0), and let R be rotation about the origin by 1 radian; then let A be the set of points you can get from (0,0) by applying some sequence of T and R, let B consist of the points where the last operation in the sequence was T rather than R, and let C be everything else. Then B = T(A) and C = R(A). (The key point here is that it turns out that the sequence of Ts and Rs is unique, so that B is well-defined.)

It is the theorem from a dead-tree math-textbook where it is used as a step in a Banach-Tarski Paradox proving.

The textbook is not in English. Here's a translation of the theorem:

An orbit O [1] can be decomposed into 4 sets: A, B, C, D. Using rotation these sets can be combined into 2 orbits:

  A ∪ aB = O; C ∪ bD = O

Proof:

  A = H(a)x; B = H(a')x; C = H(b)x; D = H(b')x

The theorem statement follows from the fact that the free group H can be decomposed into 4 parts:

  H(a),H(a'),H(b),H(b')

and doubled by rotations:

  H = aH(a') ∪ H(a); H = bH(b') ∪ H(b)

∎

__ [1]: The term `orbit` is used in the same sense as in [2]. `H` is a free group similar to the one from the step 3 in [2], and `a`, `b` are rotations defined similar to Step 2 in [2] i.e., they are generators of H:

  H = {e}∪H(a)∪H(a')∪H(b)∪H(b')

, where `e` is the unit of the group H:

  aa' = e; bb' = e

__ [2]: http://en.wikipedia.org/wiki/Banach–Tarski_paradox