For each integer $n>1$ let $F_n=\left\{\frac{k}n:k=0,\dots,n-1\right\}\;;$
for $0\le a, $|(a,b)\cap F_n|\ge\lceil n(b-a)\rceil-1$.
The first $2$ terms of the sequence are the members of $F_2$; the next $3$ terms are the members of $F_3$, and so on. Thus, the first $\sum_{k=2}^nk=\frac12n(n+1)-1$ terms are the members of blocks $F_2$ through $F_n$ in the obvious order. For $n\ge 2$ let $T_n=\frac12n(n+1)-1$, and suppose that $T_m\le N. Then
$\begin{align*} |\{n\le N:\xi_n\in(a,b)\}&\ge\sum_{k=2}^m|(a,b)\cap F_k|\\ &\ge\sum_{k=2}^m\Big(\lceil k(b-a)\rceil-1\Big)\\ &=\sum_{k=2}^m\lceil k(b-a)\rceil-m+1\\ &\ge\sum_{k=2}^mk(b-a)-m+1\\ &=(b-a)T_m-m+1\;, \end{align*}$
so
$\frac{|\{n\le N:\xi_n\in(a,b)\}}N\ge(b-a)\frac{T_m}N-\frac{m-1}N\;.$
Now $\frac{T_m}N>\frac{T_m}{T_{m+1}}=\frac{\frac12m(m+1)-1}{\frac12(m+1)(m+2)-1}\to 1\quad\text{ as }\quad m\to\infty\;,$ and
$\frac{m-1}N\le\frac{m-1}{T_m}=\frac{m-1}{\frac12m(m+1)-1}\to 0\quad\text{ as }\quad m\to\infty\;,$
so $\lim_{N\to\infty}\frac{|\{n\le N:\xi_n\in(a,b)\}|}N\ge b-a\;.$