No, let $X$ be the set of those irrationals in $x\in (0,1)$ with binary expansion
$$x=0.x_1x_2\dots$$
such that if we define $x^{\text{even}}, x^{\text{odd}}$ by
$$x^{\text{even}}=0.x_2x_4x_6\dots$$
$$x^{\text{odd}}=0.x_1x_3x_5\dots$$
then
exactly one of $x^{\text{even}}$, $x^{\text{odd}}$ is irrational.