TITLE: A non-trivial property of all groups
QUESTION [24 upvotes]: This question appeared in my answer to this  question, but it seems to be interesting in itself. Let $G$ be an infinite finitely generated group, $\epsilon\gt 0$. Is there a finite subset $S\subset G$ such that every subset of $S$ with at least $\epsilon|S|$ elements generates $G$? If the answer is "yes", it should have a trivial proof by Gromov's thesis (every property of all finitely generated groups is either false or trivial). 
 Update.  In view of Stephen's answer and Kevin's comment below, perhaps a more correct question is this: 

Is it true that if we represent an infinite group $G$ as a union of a finite number of subsets, then one of these subsets generates a finite index subgroup of $G$? 

Compare with Adreas Thom's question.

REPLY [13 votes]: This concerns the updated question. The answer is yes.
If $G$ is represented as the union of a finite number of subsets $A_1, \dots,A_n$, then one of the subsets generates a finite index subgroup of $G$. Indeed, let $G_i$ be the subgroup which is generated by $A_i$. Then, $G$ is the union of the $G_i$.
B.H. Neumann proved that if $G$ is the union of a finite number of left cosets of subgroups, then one of the subgroups is of finite index (see this post). In particular, there exists some $i$ such that $G_i$ is of finite index.