Here is how to contrive the set in plain English.
Start with the set of all points in the plane with rational coordinates, call it $S$. No translation of $S$ intersects $S$ unless it is $S$.
Let $C$ denote the set of all translations of $S$, including $S$ itself.
For each element $S^\prime$ of $C$ pick one point of $S^\prime$ in a unit square $M$ and let the set of all those points be your choice set $E$.
If $E$ is translated left, right, up or down by a rational amount, the translated set cannot intersect $E$.
Let $G$ denote the set of all translations of $E$ left, right, up or down by an amount equal to a rational number in the interval $[-1,1]$
Then $G$ is a countable collection of mutually exclusive sets which covers the unit square $M$ but whose union lies within a $3\times3$ square centered at $M$.
I assume you already know how this contradicts any assumption of $E$ having positive measure or measure $0$.