Given an $R$-module $M$ ($R$ is commutative) and a $S\subseteq {\rm Ass}_R(M)$, I want to show that there is a sub-module $N$ of $M$ s.t.${\rm Ass}_R(N)=S$.
So far, I've used Zorn's Lemma on the set of sub-modules $N'$ such that ${\rm Ass}_R(N')\subseteq S$, to show that this set contains a maximal element $T$. I don't see why ${\rm Ass}_R(T)$ could not be a proper subset of $S$.
I presume that $R$ is commutative with unity.
I will use the following theorem:
Suppose $\as(T)\subsetneq S$. Then there is a prime ideal $P\in S\setminus\as(T)$ such that $P=\newcommand{\an}{\text{Ann}}\an(x)$ for some $x\in M\setminus T$. Consider the submodule $T+Rx$. Then $$\as(T+Rx)\subseteq \as(T)\cup\as((T+Rx)/T)=\as(T)\cup\as(Rx/(T\cap Rx)).$$ If $\as(Rx/(T\cap Rx))=\emptyset$, then $\as(T+Rx)\subseteq S$, a contradiction to the maximality of $T$. So assume $\as(Rx/(T\cap Rx))\neq \emptyset$. Let $Q=\an(\bar y)$ be a prime ideal for some $y\in Rx$. Clearly, $P\subseteq Q$. Is $Q\subseteq P$? Yes!
Suppose $a\in Q$ but not in $P$. Then $a\bar y=0\neq ax$. Let $y=rx$. Then $a\bar y=0$ implies $arx\in Rx\cap T$. In particular, $arx\in T$.
I claim that $\an(x)=\an(arx)$.
Clearly $\an(x)\subseteq\an(arx)$. Now, let $z\in \an(arx)$. So $zarx=0$ and therefore $zar\in\an(x)=P$. Since $P$ is prime and $a\notin P$, we have $zr\in P$. So either $z\in P$ or $r\in P$; but if $r$ were in $P$, then that would imply $y=0$ and hence $Q=(1)$, a contradiction.
So $\an(x)=\an(arx)$. But $arx\in T$. Therefore $P=\an(x)\in\as(T)$, a contradiction to the choice of $P$. So $a\in P$ and hence $Q\subseteq P$.
So, we have $P=Q$. Therefore $\as(T+Rx)\subseteq \as(T)\cup\{P\}\subseteq S$, again a contradiction.