Contraction of square of ideal is square of contraction - unramified primes of splitting field

Question

The question is about the answer to this question. Basically we have a ring extension $R\subset S$ where $S/R$ is finite (quotient as abelian groups), $S$ is a Dedekind domain and an ideal $P$ of $S$. It is claimed that if $P$ is divisible by the square $Q^2$ of a prime ideal $Q$ then also $p:=P\cap R$ is divisible by a square of an ideal ($=q:=Q\cap R$) in $R$. To get the contradiction which will prove the original statement of the question it would suffice to prove that $p$ is just contained in $q^2$ (which is weaker in this case as $R$ might not be a Dedekind domain). Note that we have $p=P\cap R\subset Q^2\cap R$. However it is not true in general (for arbitrary rings) that $Q^2\cap R\subset (Q\cap R)^2$. Does this hold in the special case of this answer or is there maybe another way to prove the original question that would avoid this?
Edit: I found another answer to the original problem, here. This should also generalize to the case where the base field is not $\Bbb Q$, but I wonder if the approach above in the other answer would also work.