Let $x,y\in \mathbb R$. Let $x\sim y:=x-y\in \mathbb Q$. Consider the equivalence class of $x$, $[x]:=\{y\in\mathbb R\mid x\sim y\}$. By axiom of choice there exists $V\subset [0,1] $ with $\mid V\cap [x]\mid=1$, i.e. there is exactly one representative of $[x]$ in $V$. Let $q_1,q_2,\dots$ an enumeration of $\mathbb Q\cap [-1,1]$.
Then $V_k:=\{v+q_k\mid v\in V\}$ is pairwise disjoint for each $k$. That is what I do not understand. Why is this true?
To finish this argument we get $[0,1]\subset\bigcup V_k \subset [-1,2]$. By usage of $\sigma$ additivity and invariance under translation of measur $\mu$ we get $1\le\sum_k \mu(V)\le 3$ what is a contradiction.
Are the only assumptions about this measure $\mu$ that it is invariant under translation and $\mu([0,1])=1$?
Suppose that $V_k$ and $V_h$ have nonempty intersection, and let $a \in V_k \cap V_h$.
Then there exist $v,w \in V$ such that $$a=v+q_k$$ $$a=w+q_h$$ But now, notice that $$v-w = (a-q_k)-(a-q_h)=q_h-q_k \in \Bbb Q$$ thus $v \sim w$. Since $v,w \in V$, necessarily $v=w$. Then $q_h=q_k$, and from this we deduce $h=k$.
This means that $V_h$ intersects $V-K$ only when they are equal.
As for the second question: there is another assumption here to get the contradiction:
This tells you that it's impossible to define a $\sigma$-additive measure on $\Bbb R$ which is translation inveriant such that $\mu ([0,1])=1$ and measures all sets.