Let $I$ be an interval: $I=(0,1)$ Define $R$ to be relation on $I$ such that: $$(x,y)\in R\Leftrightarrow|x-y|\in\Bbb{Q}$$ It is straightforward to show that $R$ is an equivalence relation. Thus we can split $I$ into equivalence classes. The first class contains all rationals and then in each class there is some irrational number $\xi$ along with numbers which differ from $\xi$ by a rational. We can say that each equivalence class will correspond to one irrational number from $I$ (except from the first one). Let's invoke the axiom of choice to construct a set $S$, such that it contains an element from each class. $s_1$ from the first class, $s_2$ from the second ... and so on. We have: $$S=\{s_1,s_2,s_3\}$$ Now, let $r_1,r_2,...$ be a list of all rational numbers in $(-1,1)$. Now define family of sets $S_k$ for $k\in \Bbb{N}$: $$S_k=\{s_1+r_k,s_2+r_k,s_3+r_k,\cdots\}$$ so $S_k$ is just $S$ shifted by $r_k$. Now there are two big statements to consider: $$\forall i\neq j:S_i\cap S_j=\emptyset\tag{1}$$ $$\text{Every number from $I$ is in exactly one of $S_k$}\tag{2}$$ Let $\mu$ denote the Lebesgue measure. By (1) we have $$\mu\bigg(\bigcup_i S_i\bigg)=\sum_i\mu(S_i)$$ We also know that $\bigcup S_i\subseteq (-1,2)$ thus $\mu(\bigcup S_i)\leq 3$ and by (2) we have $I\subseteq \bigcup S_i$ thus $\mu(\bigcup S_i)\geq 1$.
Because $S_k$'s are just shifted $S$'s, we have $\mu(S)=\mu(S_1)=\mu(S_2)\cdots$ thus $$\mu\bigg(\bigcup S_i\bigg)=\sum \mu(S)\tag{a}$$ (The right hand side of (a) is an infinite sum) Which is impossible, because if $\mu(S)=0$ then $\mu(\bigcup S_i)=0$ and if $\mu(S)>0$ then $\mu(\bigcup S_i)=\infty$ which shows $S$ is non-measurable.
Now, I'd be interested in proving the facts (1) and (2). (1) somehow makes sense to me, because there is no way how we can "jump" from rational number to irrational, because $\Bbb{Q}$ is closed under addition and subtraction. For (2), that would also make sense, because we know that each equivalence class except the first one corresponds to one irrational number in $I$ thus that will be some $s_m+r_k$ for $m\neq 1$ which we will find in the $S_k$. For any rational, we will find it in some $S_l$ and it will correspond to $s_1+r_l$. Is my reasoning correct? I'd like some rigorous proof for these two lemmas.