The original question was:
If $A \subseteq \mathbb{R}$ and for every positive $ n \in \mathbb{N}$, such that $\vert A \cap (-n, n) \vert = \aleph_0$, then $\vert A \cap \mathbb{Q} \vert \neq \varnothing$.
I Think the claim is wrong, but I'm not sure I found a counterexapmle.
My idea for a counterexapmle is to look at the smallest interval possible, that is $(-1, 1)$, and to claim that $A = (0, \frac{1}{2}) \cap \mathbb{Q} \subseteq (-1, 1)$. Obviously $\vert A \vert = \aleph_0$. Then, for every $a \in A$, we'll add an irrational number, say $\pi – e$, and name the new subset $B$. We'll notice that $B$ still satisfies $B \subseteq (-1, 1)$. It can be shown that there exist a function that is a bijection $g: A \to B$, hence $\vert B \vert = \aleph_0$ and $\vert B \cap \mathbb{Q} \vert = \varnothing$.
Any help would be great.
Best Answer
You have the right idea, and you can do the adding an irrational number to everything in $A$ at once. That is, define $$ A = \{q + \sqrt{2} : q \in \mathbb{Q}\}. $$ Then $A \cap \mathbb{Q} = \emptyset$, while $|A \cap (-n, n)| = \aleph_0$ for every $n > 0$.