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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.