How to contruct such a sequence of rational? How to order all  rational numbers from $(0,1)$ in a sequence $(x_n)_{n=1}^\infty$ in such a way that $$|x_n-x_k| \geq \frac{1}{(n+1)^2}$$ for $k<n$  ?
 A: Order by ascending denominator, then by ascending numerator, i.e. $$ \frac12,\quad \frac13,\frac23,\quad\frac14,\frac34,\quad\frac15,\frac25,\frac35,\frac45,\quad\frac16,\frac56,\quad\frac17,\frac27,\frac37,\ldots$$
By this, the denominator of $x_n$ is $\le n+1$ because for each denominator $d$ we have at least one entry $\frac1d$ in this list. Hence for $k<n$ we have $x_k=\frac ab$, $x_n=\frac cd$ with $b\le k+1\le n$ and $d\le n+1$, hence $$|x_n-x_k|=\frac{|\overbrace{bc-ad}^{\ne0}|}{bd}\ge \frac1{bd}\ge\frac1{n(n+1)}>\frac1{(n+1)^2}.$$

Note that for denominators $d>2$, we have at least two entries $\frac1d$ and $\frac{d-1}d$ in the list. Hence the denominator of $x_n$ is in fact $\le \frac {n}2+2$. This ultimately gives us the slight improvement (for $n>2$)
$$ |x_n-x_k|>\frac{4}{(n+4)^2}.$$
A: It seems that $\dfrac{1}{(n+1)^2}$ can be improved. If we do the first thing that comes to our mind:
$$
\frac{1}{2},\,\,\,\frac{1}{3},\frac{2}{3},\,\,\,\frac{1}{4},\frac{3}{4},\,\,\,
\frac{1}{5},\frac{2}{5},\frac{3}{5},\frac{4}{5},\,\,\,
\frac{1}{6},\frac{5}{6},\,\,\,\frac{1}{7},\frac{2}{7},\frac{3}{7},\frac{4}{7},
\frac{5}{7},\frac{6}{7},\ldots
$$
then $x_n$ is a fraction with denominator at $n$, and if $k<n$, then $x_k$ has numerator either $n+1$ or less than $n+1$. If its numerator is $n+1$, then $|x_k-x_n|\ge\frac{1}{n+1}$, whereas if it is less than $n$, then
$$
|x_n-x_k|\ge\frac{1}{n(n+1)}.
$$ 
