$A \subset \mathbb{R}$ uncountable $\Rightarrow$ $ \exists ~I$ interval such that $A \cap I$ is dense in $I$? If $A \subset \mathbb{R}$ is uncountable, does it follow that there exists a non-degenerate interval $I$ such that $A \cap I$ is dense in $I$?
I was thinking that Cantor set might be a counterexample, but eventually I reached the conclusion that it isn't. I'm not really sure, though.
I think that we can assume WLOG that $A \subset [0,1]$. This is because if we choose $A_k=A \cap [k,k+1]$, then $A= \bigcup\limits_{k \in \mathbb{Z}}A_k$, and since this is a countable union, it follows that there exists $n \in \mathbb{Z}$ such that $A_n$ is uncountable. We can assume WLOG that $n=0$ and then we can just take the problem from the beginning, but now with $A_0$ instead of $A$.
I tried to solve the problem by contradiction, but I didn't succeed.
 A: What you are asking is whether for an uncountable set $A$ the closure of $A$ has a non-void interior. So any Cantor set (uncountable, closed, and not containing an interval) will be a counterexample. Note that the Cantor space does not have connected subsets with more than one point, so this will work in many other spaces.
A: As said in the comments, the cantor set is a counterexample, as it has empty interior, so it is nowhere dense (if $C\cap I$ was dense, then it would equal $I$, since $C$ is closed).
Similar examples show that even if you strengthen the hypothesis to say that $A$ is of positive Lebesgue measure, this is still not true. However in this case, you do get something: by the Lebesgue density theorem, you have that if $A$ has positive measure (in fact, as large as you want, up to but not including full measure), then $A$ is dense at almost every point of $A$ in the sense of approximate density.
This shows that having large cardinality and measure imply nothing about category (in the sense of Baire category).
