Open superset of $\mathbb{Q}$ Let $S$ be an open set such that $\mathbb{Q}\subset S$. We can also define a set $T=\mathbb{R}\setminus S$. I have been trying to prove or disprove whether $T$ could be uncountable. I suspect $T$ has to be at most countable, is my intuition correct?

Your intuition is reasonable, but incorrect. HINT: the rationals are countable, so we can list them as $\mathbb{Q}=\{q_1, q_2, q_3, . . .\}$ (obviously this listing isn't "in order," in any sense, but that's fine). 
Now, let $U_n=(q_n-2^{-n}, q_n+2^{-n})$. Each $U_n$ is open, so the union $V=\bigcup U_n$ is open, and clearly $V$ contains $\mathbb{Q}$. Do you see how to show that $\mathbb{R}\setminus V$ is uncountable?

If you are unfamiliar with Lebesgue measure, the above hint will be very difficult. An alternate approach is via the Baire category theorem: let $C$ be the Cantor set, and show that there is some real $r$ such that $C+r=\{x+r: x\in C\}$ contains no rational numbers. Then the complement of $C+r$ is open, contains the rationals, and has uncountable complement. HINT: To use BCT, show that for each rational $q$, the set $B_q=\{r: q\in C+r\}$ is nowhere dense. Then, since there are only countably many rationals, BCT implies that $\bigcup_{q\in\mathbb{Q}} B_q$ is meager, and hence $\not=\mathbb{R}$ . . .
