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Forcing a symmetric sequence in $[0,1]$

Simple question at a pre-graduated level:

Given a sequence $(x_n)$ taking values in the interval $[0,1]$, is it always possible to construct a "symmetrizing" sequence $(s(n))$ of $0$ or $1$, such that the sequence $$ y_n \doteq (-1)^{s(n)}x_n + s(n) $$ is "symmetric in $[0,1]$" in the sense that for any $ 0 \leq a < b < \frac{1}{2} $, the sequence $$ \text{Card} ( n \leq N : y_n \in [a,b] ) - \text{Card} ( n \leq N : y_n \in [1-b,1-a]) $$ is bounded as $N$ increases?

I think I prove the result for $|b-a|>\epsilon$ for any $\epsilon > 0$ (we could say $\epsilon-$symmetric).

But I wonder if it generalizes to $\epsilon = 0$ a.k.a for $\Sigma_\epsilon \doteq \left( s:s \text{ } \epsilon -\text{symmetrises } x \right)$, if we have $$ \bigcap_{\epsilon > 0} \Sigma_\epsilon \neq \emptyset \text{ ?} $$ which is the existence of a symmetrizer $s$ of the sequence $x$.

PS : My final goal would be to know (if such symmetrizer $s$ exists) if for $x$ sequence, the symmetrized $y$ necessarily admits an historical density in $[0,1]$ talking about the Weyl meaning of density.