Motivation: Let $G$ be any group such that all its subgroups are finitely generated, then it is easy to see that $G$ satisfies ACC.
We know there are finitely generated groups that don't satisfy the above, for example see this answer. So my question is can we have an example of a finitely generated group which does not satisfy ACC?
If $G$ is a free group in two generators, there is a subgroup $F$ of $G$ which is isomorphic to a free group in countably many generators; for example, we could take $F=G'$ the derived subgroup of $G$.
Suppose the generators are $\{x_n:n\in\mathbb N\}$. Pick any bijection $\phi:\mathbb N\to\mathbb Q$, and for each real number $r\in\mathbb R$ let $H_r$ be the subgroup of $G$ generated by the set $\{x_n:\phi(n)<r\}$.This provides us with an uncountable set $\{H_r:r\in\mathbb R\}$ of proper subgroups of $G$ such that $$r<s\implies H_r\subsetneq H_s.$$ You can easily find now an ascending chain of subgroups of $G$ which does not stabilize!