No, let $X$ be the of those irrationals in $x\in (0,1)$ with binary expansion $$x=0.x_1x_2\dots$$ such that if we use a bijection $\mathbb N^2\to\mathbb N$ to represent $x$ as infinitely many reals $x^{(1)}, x^{(2)},\dots$ then $$\lim_{n\to\infty} 1_{\mathbb Q}(x^{(n)})$$ exists.

Here $1_{\mathbb Q}(y)=1$ if $y\in\mathbb Q$ and 0 otherwise.

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.