I was thinking the following thing:
Is there an uncountable group whose all proper subgroups are countable which is also for instance locally soluble?
I've found some example of minimal non-countable groups which are perfect or Artinian, but nothing in the locally soluble case (or some other good generalized soluble condition). Is it possible maybe to prove that they do not exist in the locally soluble case for instance?
No. It is not so difficult to prove it when $G$ is abelian. So assume that $Z(G)$ is a proper subgroup of $G$. Let $N$ be a countable normal subgroup of $G$ not contained in the center. If we take an element in $N$ not in the center, then it has uncountable many conjugates in $G$, and hence the subgroup generated is uncountable, hence $N$ is uncountable. Thus $G/Z(G)$ is simple.