Is any uncountable, scattered subset of [0,1] dense? Just out of curiosity, I am wondering if any uncountable, scattered subset $U\subset[0,1]$ must be dense in $[0,1]$ (endowed with the Euclidean topology). It's not necessarily true if $U$ is countable, since we can take $U=\{\frac{1}{n} : n\in\mathbb{N}\}$, and there are countably many open sets in $[0,1]$ which $U$ does not intersect.

But I'm having a hard time figuring out whether it's true for uncountable $U$. Any help would be appreciated. (I added the "scattered" hypothesis) since any other interval contained in $[0,1]$ would suffice as a counterexample).
There are no uncountable scattered subsets of $[0,1]$.  This follows from the theory of Cantor-Bendixson rank, for instance.  Given any scattered $A\subset[0,1]$, there must be some ordinal $\alpha$ such that the $\alpha$th Cantor-Bendixson derivative $A^\alpha$ of $A$ is empty.  The least such $\alpha$ must be countable, since the Cantor-Bendixson derivatives are a descending chain of closed subsets of $A$ and $A$ is second-countable.  Since every subset of $A$ can have only countably many isolated points (again by second-countability), this means $A$ is countable.
An uncountable set does not have to be dense. A simple example would be to take the half interval $[0,\frac{1}{2}]$ which is uncountable but not dense in $[0,1]$. But we can do even better, because we can find a set that is uncountable and not dense in any open interval. The Cantor set is uncountable and dense nowhere. https://en.wikipedia.org/wiki/Cantor_set