I recently asked this question about whether or not profinite groups admit maximal subgroups: And indeed, profinite groups admit subgroups of finite index, so taking any minimum index subgroup containing the finite index subgroup gives a maximal subgroup of the profinite group.
Now I wonder if it's true that all subgroups of a profinite group are contained in a maximal subgroup, i.e., can you have infinite chains of infinite index subgroups which do not split off to maximal subgroups periodically as you go up the chain?
Consider the integers as an additive subgroup of the $p$-adic integers: $\mathbb{Z}\subset\mathbb{Z}_p$ is not contained in its maximal subgroup $M=p\mathbb{Z}_p$. Let's show this maximal subgroup is unique:
Suppose $M<\mathbb{Z}_p$ is maximal, so $\mathbb{Z}_p/M=\langle x+M\rangle\cong\mathbb{Z}/q\mathbb{Z}$ for some prime $q$. If $q\ne p$ then $\frac{1}{q}x+M$ would have order $q^2$ in the quotient, a contradiction, so $q=p$. This implies $p\mathbb{Z}_p\subseteq M$, which forces $M=p\mathbb{Z}_p$ by transitivity of index.