If a finite translation of $A$ covers an abelian group, infinite translations of it intersect. Let $G$ be an abelian group and $A\subseteq G$. Suppose there's a finite set $F\subseteq G$ such that:
$$G=FA$$
How can I prove any infinite translation of $A$ is overlapping, that is, there's not any infinite set $C\subseteq G$, such that for each $x,y\in C$,
$$x\ne y \quad \leftrightarrow\quad xA \cap yA\ne \emptyset$$
?


Edit:
  (according to comments below)
  More generally we can say if for some $A_1,...,A_n\subseteq G$,
  $$A=\bigcup_{k=1}^nA_k$$  then there's  some $k$ such that any infinite translation of $A_k$ is overlapping.

The proof suggested below includes existence of a Banach measure. I wonder if there's any elementary proof for this.
 A: I'll switch to additive notation, because this problem becomes easier if we remind ourselves that $G$ is abelian as often as possible.
Suppose that $G=F+A$, where $G$ is abelian and $F$ is finite. Suppose $|F'|>|F|$ and consider the translates of $A$ by $F'$. I claim they overlap somewhere.
Since $-F'\subset G = F+A$, by the pigeonhole principle there exists $f_1,f_2\in F'$ such that $-f_1,-f_2\in f+ A$ for the same $f\in F$, so $-f\in (f_1+A)\cap(f_2+A)$.

Adding the observation of Alex Ravsky, the proposition clearly extends to the class of all amenable groups, which includes in particular all solvable groups. The proof in this general case is quite obvious and intuitive: if $\mu$ is a left-invariant mean on $G$, and $A$ has $n$ pairwise disjoint translates, then $\mu(A)\leq \frac{1}{n}$, so certainly no fewer than $n$ translates of $A$ suffice to cover $G$.
With this perspective we can answer OP's harder question affirmatively: Given a subset $A=\cup_{k=1}^n A_k$ of an amenable group $G$ and a finite set $F$ such that $G=FA$, there exists $k$ such that no more than $n|F|$ left-translates of $A_k$ can be pairwise disjoint. Indeed, $\mu(A)\geq\frac{1}{|F|}$, so for some $k$ we must have $\mu(A_k)\geq \frac{1}{n|F|}$, and then clearly any $n|F|+1$ translates of $A_k$ are not disjoint.
On the other hand, the claim does not hold in the class of all groups, as the group $G=\mathbf{Z}\ast C_2 = \langle x,y\,|\,y^2=e\rangle$ shows: Take $A$ to be all words beginning with $y$. Then $G=A\cup yA$, but $A,xA,x^2A,\ldots$ are all disjoint.
