 Solution

Take the middle-thirds Cantor set $C$. Its $n$th stage is the union of $2^n$ closed intervals of length $3^{-n}$, and contains $C$. Their total length is $(2/3{)}^n\to 0$; slightly enlarging them to open intervals gives covers of arbitrarily small total length. Therefore $C$ is null.

For uncountability map every binary sequence $({\epsilon }_j{)}_{j\ge 1}$ to ${\sum }_{j\ge 1}2{\epsilon }_j{3}^{-j}\in C$. If two sequences first differ at index $k$, their image difference in absolute value is at least $2\cdot 3^{-k}-{\sum }_{j>k}2\cdot 3^{-j}=3^{-k}>0$. Thus the map is injective. The set of binary sequences is uncountable by Cantor's diagonal argument, so $C$ is uncountable.