Open cover rationals proper subset of R? If I were to cover each rational number by a non-empty open interval, would their union always be R? It seems correct to me intuitively, but I am quite certain it is wrong. Thanks

Take an irrational number $x \in \mathbb{R} - \mathbb{Q}$. If $q \in \mathbb{Q}$ is any rational number then $\left| x - q \right| = r_q > 0$. The union $\displaystyle \bigcup_{q \in \mathbb{Q}} (q-r_q, q+r_q)$ then does not contain $x$, and so particular does not cover $\mathbb{R}$.
[Interestingly, it does cover $\mathbb{R}-\{x\}$; but you could change your intervals in the union to force this not to be the case.]
