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3 typo

Let S be a subset of the reals such that S∩[a,b] and Sc∩[a,b] cannot be written as a countable union of closed sets for any a<b. This can be done (this explicit example of a non-Borel set achieves this). Let ℚ be the rationals. Then, A=(Sxℚ)U(Scxℚc) and B=(Sxℚc)U(Scxℚ) 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-1(ℚ))=∪x∈ℚf(I∩g-1(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 into either ℚ or ℚc, so is also constant. So A is totally path disconnected. The argument for B follows in the same way by exchanging S and Sc