Let $A \subset \mathbb{R}^2$ be countable. Then it is not too hard to show that $\mathbb{R}^2 \setminus A$ is path-connected. However it is not always Manhattan connected since if $A = \mathbb{Q}^2 \setminus \{(0,0)\}$, the origin cannot be connected to any other point by a path moving only horizontally or vertically.
Say a subset of $\mathbb{R}^2$ is skew-Manhattan connected if there exist two directions so that any two points in the set can be connected by a path which moves in only those two directions.
What I want to decide is whether $\mathbb{R}^2 \setminus A$ as above is always skew-Manhattan connected?