Let $d(X)$ be the asymptotic density of a set $X$ of positive integers, i.e., $$ d(X):=\lim_{n\to \infty}\frac{|X\cap [1,n]|}{n}, $$ whenever this limit exists.
Fix also a sequence $(x_n)$ of reals taking values in the closed interval $[0,1]$.
Question. Is it possible that, for each real $x$, there exists an open neighborhood $I$ of $x$ such that $ d(\{n: x_n \in I\})=0? $
No, it's not possible.
Since $[0,1]$ is compact, we find a finite set of neighborhoods $U_1, \ldots, U_m$ covering $[0,1]$.
Assume that $d(\{n:x_n \in U_k\})$ exists and equals zero for every $k$.
For any $\epsilon > 0$ we have an $n$ such that for every $k$ we have $$ \frac{|\{i\leq n: x_i \in U_k|}{n} < \epsilon. $$
It follows $$ 1 \leq \sum_{k=1}^m \frac{|\{i\leq n: x_i \in U_k|}{n} < m\epsilon $$
where the first inequality is due to the $U_k$ covering $[0,1]$.
by picking $\epsilon$ small enough, in particular less than $\tfrac{1}{m}$ we have a contradiction.