Why is this set uncountable? Small covers of rationals in $[0,1]$. 
Let $\{q_n : n\in\mathbb{N} \}$ an enumeration of $\mathbb{Q}\cap[0,1]$. For any $\varepsilon > 0$ set
  $$A(\varepsilon) = \bigcup_{n=1}^\infty \left(q_n-\frac{\varepsilon}{2^n}, q_n+\frac{\varepsilon}{2^n} \right) $$
  and let
  $$A = \bigcap_{k=1}^\infty A\left(\frac 1 k \right).$$

I can show that it's uncountable (use Baire's) and has measure zero but my gut really says it should be countable because what else could be in this intersection other than rationals?
 A: Each $A(\epsilon)$ is open and dense; this means that we can inductively find a sequence $I_n$ of nonempty open intervals such that


*

*The closure of $I_{n+1}$ is contained in $I_n$.

*Each $I_n\subseteq A({1\over n})$.

*$q_n\not\in I_n$.
By the first condition, $J:=\bigcap_{n\in\mathbb{N}} I_n\not=\emptyset$. Let $x\in J$; then by the second condition, $x\in A({1\over n})$ for each $n$, hence $x\in A$. And by the third condition, $x\not\in\mathbb{Q}$.

Thinking along these lines will eventually lead you to the Baire category theorem, which among other things implies that any subset of $\mathbb{R}$ which is dense and a countable intersection of open sets (such sets are variously denoted $G_\delta$ or $\Pi^0_2$) is not a countable union of nowhere dense sets; since every countable set is a countable union of nowhere dense sets (namely, singletons), this means that every dense $G_\delta$ set is uncountable,$^*$ and your $A$ is clearly dense $G_\delta$.
$^*$More is true: we can in fact show that any uncountable $G_\delta$ set, and indeed any uncountable analytic set, has size continuum.
