As I mentioned in a comment above there is a nice set $X$ containing half the rationals with a very simple description: the set of reduced fractions $a/b$ with $\operatorname{gcd}(ab,3)=1$.

To see it, we can map each reduced fraction $a/b$ to the pair $(x,y) \equiv (a,b) \pmod{3}$. There are 8 possible pairs $(x,y)$:
$$ (0,1),(0,2),(1,0),(2,0),(1,1),(1,2),(2,1),(2,2) $$
if we can prove that reduced fractions in an interval $Y$ are uniformly distributed between these 8 pairs then we have the result.

It is clearly enough to prove this for an interval $Y = [0,\lambda)$ for some real $\lambda > 0$. Lets fix a pair $(x,y)$ and call $A_{(x,y)}(n)$ the number of fractions $a/b$ with $b\le n$, $\operatorname{gcd}(a,b)=1$, $a < \lambda b$, and $(a,b)\equiv(x,y\pmod{3}$ ie the horrid sum;

$$ A_{(x,y)}(n) = \sum_{b\le n}\sum_{\begin{matrix}\operatorname{gcd}(a,b)=1\\(a,b)\equiv (x,y) \pmod{3}\\a \le \lambda b\end{matrix}} 1 $$

Now we use the Möbius identity:
$$
\sum_{d \vert \operatorname{gcd}(a,b)}\mu(d) = \begin{cases}1 \quad&\text{if }\operatorname{gcd}(a,b)=1\\0&\text{otherwise}\end{cases}
$$
to get

$$ A_{(x,y)}(n) = \sum_{b\le n}\sum_{\begin{matrix}(a,b)\equiv (x,y) \pmod{3}\\a \le \lambda b\end{matrix}} \sum_{d\vert \operatorname{gcd}(a,b)} \mu(d) $$

We can invert this sums puting $b = kd, a = td$ and then $t < \lambda k$ so we have

$$ A_{(x,y)}(n) = \sum_{d\le n} \mu(d) \sum_{\begin{matrix}kd\le n\\kd\equiv y\pmod{3}\end{matrix}}\sum_{\begin{matrix}t<\lambda k \\td\equiv x\pmod{3}\end{matrix}} 1 $$

In the outer sum we can limit ouselves to integers $d$ coprime with 3, (if $3 \vert d$ as $(x,y)\not\equiv (0,0)\pmod{3}$ there is nothing to sum) so we have:

$$
\begin{align} A_{(x,y)}(n) &= \sum_{\begin{matrix}d\le n\\d\not\equiv 0\pmod{3}\end{matrix}} \mu(d) \sum_{\begin{matrix}kd\le n\\kd\equiv y\pmod{3}\end{matrix}}\sum_{\begin{matrix}t<\lambda k \\td\equiv x\pmod{3}\end{matrix}} 1\\
&=\sum_{\begin{matrix}d\le n\\d\not\equiv 0\pmod{3}\end{matrix}} \mu(d) \sum_{\begin{matrix}kd\le n\\kd\equiv y\pmod{3}\end{matrix}} \left(\frac{\lambda k}3 + O(1)\right) \\
&=\sum_{\begin{matrix}d\le n\\d\not\equiv 0\pmod{3}\end{matrix}}
\mu(d) \left( \frac{ \lambda n^2}{18d^2} + O(n/d) \right)\\
&=\frac{\lambda n^2}{18}\sum_{\begin{matrix}d=1\\d\not\equiv 0\pmod{3}\end{matrix}}^\infty \frac{\mu(d)}{d^2} + O(n) + O(n\log n)
\end{align}
$$

We can evaluate the constant in the last equation, call
$$ B = \sum_{\begin{matrix}d=1\\d\not\equiv 0\pmod{3}\end{matrix}}^\infty \frac{\mu(d)}{d^2} $$
then
$$ B - \frac{B}{9} = \sum_{\begin{matrix}d=1\\d\not\equiv 0\pmod{3}\end{matrix}}^\infty \frac{\mu(d)}{d^2} +\sum_{\begin{matrix}d=1\\d\not\equiv 0\pmod{3}\end{matrix}}^\infty \frac{\mu(3d)}{(3d)^2} = \sum_{d=1}^\infty \frac{\mu(d)}{d^2} = \frac{1}{\zeta(2)} = \frac{6}{\pi^2}$$
and so:
$$ B = \frac{27}{4\pi^2} $$
and finally
$$ A_{(x,y)}(n) = \frac{3\lambda }{8\pi^2} n^2 + O(n\log n) $$
so each pair $(x,y)$ have ultimately the same proportion $1/8$ of all reduced fraction in the interval, as was to be shown.

Note: I suppose that with some additional work it can be proven that for a given modulus $m$ the reduced fractions $a/b$ are uniformly distributed when reducing mod $m$ between the pairs $(x,y)$ such that $\operatorname{gcd}(x,y,m)=1$. And then we could find sets containing any given proportion of the rationals with the same procedure.