Let S be a subset of the reals such that S∩[a,b] and S<sup>c</sup>∩[a,b] cannot be written as a countable union of closed sets for any a<b. This can be done (this <a href="http://planetmath.org/?op=getobj&from=objects&id=11351">explicit example of a non-Borel set</a> achieves this). Let ℚ be the rationals. Then, A=(Sxℚ)U(S<sup>c</sup>xℚ<sup>c</sup>) and B=(Sxℚ<sup>c</sup>)U(S<sup>c</sup>xℚ) should do it. The proof is as follows. Suppose that the curve t→(f(t),g(t)) lies in A, and consider a closed bounded interval I. As the curve lies in A, f(I)∩S = f(I∩g<sup>-1</sup>(ℚ))=∪<sub>x∈ℚ</sub>f(I∩g<sup>-1</sup>(x)) is a union of countably many closed sets. By the choice of S, f(I) must be a single point. Hence, f is constant. Then, g is a continuous function mapping onto either ℚ or ℚ<sup>c</sup>, so is also constant. So A is totally path disconnected. The argument for B follows in the same way by exchanging S and S<sup>c</sup>