Consider the set $E=\{ x\in\mathbb{R}^3\mid x\notin\mathbb{Q}^3 \}$.
Is it possible to write $E$ as a union of pairwise disjoint axis parallel lines?
This seems like a problem for which the solution will probably be non-constructive, maybe using transfinite induction?
No, it is impossible. Suppose that it is possible, and consider the set of eight points whose coordinates are all either $0$ or $\pi$. (All of these points are in $E$ except for the origin.) Note that $(\pi, 0, 0)$ must belong to one of our lines, and it must be either $\{(\pi, y, 0)\}$ or $\{(\pi, 0, z)\}$, as $\{x, 0, 0\}$ is not contained in $E$. Assume without loss of generality that it is the former. Then this covers $(\pi, 0, 0)$ as well as $(\pi, \pi, 0)$. Next, look at $(0, \pi, 0)$. This could only be contained in $\{(x, \pi, 0)\}$ or $\{(0, \pi, z)\}$, but the former is impossible because it intersects our previous line at $(\pi, \pi, 0)$. A similar argument shows that $(0, 0, \pi)$ must be contained in the line $\{(x, 0, \pi)\}$ rather than $\{(0, y, \pi)\}$, because we've already hit $(0, \pi, \pi)$.
We've now found three lines, namely $\{(\pi, y, 0)\}, \{(0, \pi, z)\},$ and $\{(x, 0, \pi)\}$, that contain exactly six of our eight chosen points: all of them except $(\pi, \pi, \pi)$ and the forbidden $(0, 0, 0)$. But some line must contain $(\pi, \pi, \pi)$, and we have a contradiction: each of the three axis-parallel lines containing $(\pi, \pi, \pi)$ intersect one of the lines we've already chosen.
Edit: As Eric pointed out in the comments, the proof can be simplified vastly by observing that any axis-parallel line containing at least one of our seven chosen points in $E$ must contain exactly two of them, and seven points can't be split into pairs.