Example of an uncountable dense set with measure zero

Consider the union of $\mathbb{Q}\cap[0,1]\cup K$, where $K$ is the ternary Cantor set.

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.

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 $aX+b\subseteq I.$