If $(s_1, s_2, s_3\ldots )$ is an equidistributed sequence on $[0,1],$ then for each $\ 0<\delta<\varepsilon <1\ $ and each $\ c\in [0,1-\varepsilon],\ \exists\ N\ $ such that
$$ \varepsilon - \delta < \frac{ \left\lvert \{ s_1, \ldots, s_n \} \cap [c, c+\varepsilon] \right\rvert }{ n } < \varepsilon + \delta $$
for all $\ n\geq N.$
This implies:
If $(s_1, s_2, s_3\ldots )$ is an equidistributed sequence on $[0,1],$ then for each $\ 0<\delta<\varepsilon <1\ $ and each $\ (c_i)_{i=1}^{k}\subset [0,1-\varepsilon],\ \exists\ N\ $ such that
$$ \varepsilon - \delta < \frac{ \left\lvert \{ s_1, \ldots, s_n \} \cap [c_i, c_i+\varepsilon] \right\rvert }{ n } < \varepsilon + \delta $$
for every $\ i\in \{1,\ldots,k\}\ $ and for all $\ n\geq N.$ This is because there is an $N(i)$ for each $c_i,$ and so in the statement above, we can let $N=\max_{1\leq i\leq k} N(i).$
To me, this motivates the following question, which I assume is false, although I struggle to construct a counter-example, and am interested in how to write one.
Let $(s_1, s_2, s_3\ldots )$ be an equidistributed sequence on $[0,1].$ Is it true then, that for each $\ 0<\delta<\varepsilon <1,\ \exists\ N\ $ such that
$$ \varepsilon - \delta < \frac{ \left\lvert \{ s_1, \ldots, s_n \} \cap [c, c+\varepsilon] \right\rvert }{ n } < \varepsilon + \delta $$
for every $\ c\in [0,1-\varepsilon]\ $ and for all $\ n\geq N?$
The result is true as follows. Construct a partition $0, \delta/4,...k\delta/4, 1$ of $[0,1]$ of width $\delta/4$ except possibly for the last interval with width at most that. If $x_m, m=0,...k+1$ are the nodes of that including $1$ every point $c \in [0, 1-\epsilon]$ is within at most $\delta/4$ from a node to the left, let's call it $x_m$ so $[c, c+\epsilon] \subset [x_m, x_m+\epsilon+\delta/4]$ since $x_m \le c \le x_m+\delta/4$ and of course if $x_m +\epsilon+\delta/4 >1$ we have $[c, c+\epsilon] \subset [x_m, 1]$
But now for each $x_m$ we have $\lim_{n \to \infty} \frac{ \left\lvert \{ s_1, \ldots, s_n \} \cap [x_m, x_m+\varepsilon+\delta/4] \right\rvert }{ n }=\epsilon+\delta/4$ while in the border case the limit is $\epsilon \le 1-x_m \le \epsilon+\delta/4$
So picking $N_1$ large enough depending only on all the $x_m$, we get $\frac{ \left\lvert \{ s_1, \ldots, s_n \} \cap [c, c+\varepsilon] \right\rvert }{ n } \le \frac{ \left\lvert \{ s_1, \ldots, s_n \} \cap [x_m, x_m+\varepsilon+\delta/4] \right\rvert }{ n } < \epsilon+\delta/2$ for all $c \in [0,1-\epsilon]$ and all $n \ge N_1$
Similarly $c$ is within at most $\delta/4$ to a node to the right $x_q$ hence $[x_q, x_q+\varepsilon-\delta/4] \subset [c, c+\epsilon]$ since $c \le x_q \le c+\delta/4$ and the same reasoning give an $N_2$ depending again only on the $x_m$'s st $\frac{ \left\lvert \{ s_1, \ldots, s_n \} \cap [c, c+\varepsilon] \right\rvert }{ n } \ge \frac{ \left\lvert \{ s_1, \ldots, s_n \} \cap [x_m, x_m+\varepsilon-\delta/4] \right\rvert }{ n } > \epsilon-\delta/2$ for all $c \in [0,1-\epsilon]$ and all $n \ge N_2$ and then picking $N$ the largest of the two we are done!