As stated in the title, I am trying to find an example of an uncountable dense subset of $[0,1]$ that has measure zero. My intuition is that such a subset cannot exist, but I do not have a proof of this.
Currently, I can construct an uncountable dense subset that has arbitrarily small measure. Also, it is easy to construct an uncountable subset that has zero measure.
On
Or even (without just taking an uncountable set of measure zero and throwing in the rationals) union rational translations of the Cantor set (mod 1)
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Let $X$ be the uncountable measure zero subset of $[0,1]$ which you constructed. Let $Y$ be the union of all sets of the form $aX+b$ where $a,b$ are rational numbers, $a\gt0.$ Then $Y$ is a set of measure zero (countable union of measure zero sets) and is "uncountably dense" in the sense that every interval $I$ of the real line has uncountable intersection with $Y,$ because there are rational numbers $a,b$ with $a\gt0$ such that $aX+b\subseteq I.$
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You can even construct a set $S \subset \mathbb{R}$ such that $S \cap U$ is uncountable for every open $U \subseteq \mathbb{R}$ and still $m(S \cap U) = 0$ where $m$ is the Lebesgue measure.
To do this, we start with the Cantor set $C \subset [0, 1]$ and create "denser" sets by gluing together scaled down copies of $C$:
$$\begin{align} S_n & := \bigcup \{ 3^{-n} (x+k) \, | \, x \in C, \, k \in \mathbb Z \} \\ S & := \bigcup_{n=0}^{\infty} S_n \\ \end{align}$$
Since $S_n$ is a countable union of nullsets (sets of measure $0$), also $S_n$ will be a nullset. In the same way $S$ will be a nullset.
The numbers in $S$ will have a ternary "decimal" expansion with only a finite number of ones.
Consider the union of $\mathbb{Q}\cap[0,1]\cup K$, where $K$ is the ternary Cantor set.