Let me try. QxQ and its complement in R^2. Does this work? Edit: this does not work, :(. Let me try something else then. Take S = (R\Q)x(R\Q). QxR U RxQ would be S complement. Let p be a prime number and let q_p be in Q. For each q_p, let J_p be the set of all rational numbers times the square root of p. Since Q is countable, there is a bijection between the set of primes and {q x R : q is rational} union {R x q : q is rational}. So let A be the union of all {q_pq_p X J_p} and {J_pJ_p X q_p}. Take A union S and the complement of that in R^2 to be your disjoint union. Dunno if this works but at least it seems like a better attempt.
