Let $S$ be a subset of the reals such that $S \cap [a,b]$ and $S^c \cap [a,b]$ 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][1] achieves this). Let $\mathbb{Q}$ be the rationals. Then, $A = (S \times \mathbb{Q}) \cup (S^c \times \mathbb{Q}^c)$ and $B = (S \times \mathbb{Q}^c) \cup (S^c \times \mathbb{Q})$ should do it.

The proof is as follows. Suppose that the curve $t \to (f(t),g(t))$ lies in $A$, and consider a closed bounded interval $I$. As the curve lies in $A$, $$f(I) \cap S = f(I \cap g^{-1}(\mathbb{Q})) = \bigcup_{x \in \mathbb{Q}} f(I \cap g^{-1}(x))$$ 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 into either $\mathbb{Q}$ or $\mathbb{Q}^c$, so is also constant. So $A$ is totally path disconnected. The argument for $B$ follows in the same way by exchanging $S$ and $S^c$.

  [1]: https://planetmath.org/alebesguemeasurablebutnonborelset