Define for $\epsilon > 0 $ $$V_\epsilon = \left( \bigcup_{j \in \mathbb{N}} (x_n - \frac{\epsilon}{2^{n+1}} , x_n + \frac{\epsilon}{2^{n+1}}) \right) \ \bigcap \ (0,1)$$ where $x_n$ stems from the enumeration $\{x_1,x_2,...\}$ of the countable set of rational numbers in $[0,1]$.
Is this set dense in $(0,1)$? I figure that $V_\epsilon$ contains all rational numbers in $(0,1)$, so if $x$ is a rational in $(0,1)$ it lies in $V_\epsilon$, and if it's irrational in $(0,1)$, there's a rational (in $(0,1)$ arbitrary close to it, and this will lie in $V_\epsilon$?
No idea if this makes sense.
Yes. More generally: Any superset of a dense set is dense itself.