This is a partial answer. There exists an infinite group $G$ having no sequence of subsets $\{A_i\}$ such that 
$$
A_{n+1}A_{n+1}\subsetneq A_n,
\quad\hbox{and}\quad
\bigcap A_i=\{1\}.
$$
(We omit the inverse condition but add the the intersection-triviality condition.)

Indeed, such a sequence defines a Hausdorff topology on the group where $A_i$ are neigbourhoods of $1$ and all other points are isolated. The multiplication is continuous at $(1,1)\in G\times G$ with respect to this topology. It remains to note that infinite locally non-topologizable groups do [exist](http://mathoverflow.net/a/179233/24165).