Here $A$ is a noetherian ring and $M$ is a finitely generated $A$-module, and $m\in M$. I'm trying to solve this by assuming
(A) The associated primes/points of $M$ are precisely the generic points of irreducible components of the support of some element of $M$ (on Spec $A$).
I wrote down my proof but since I'm new to this topic, I'm not confident. So I really appreciate if anyone can take a quick look to verify:
Let $S={\{p\in \text{Supp}\ m \mid p \text{ is an associated point of }M\}}$, and we need to show that \begin{align*} \overline{S} = \text{Supp}\ m. \end{align*} Since $S \subset\text{Supp}\ m$ and Supp $m$ is closed, we have \begin{align*} \overline{S}\subset \text{Supp}\ m. \end{align*} For the other direction, note that Spec $A$ is noetherian, so Supp $m$ as a closed subset, is the finite union of its irreducible components \begin{align*} \text{Supp}\ m = \cup_{i=1}^r X_i. \end{align*}
Each irreducible component $X_i$ has a generic point $p_i$, that is \begin{align*} X_i = \overline{\{p_i\}}. \end{align*} By (A) these generic points are all associated points of $M$, so $p_i\in S$, it follows that \begin{align*} \text{Supp}\ m = \cup_{i=1}^r \overline{\{p_i\}} = \overline{\cup_{i=1}^r\{p_i\}} \subset \overline{S}. \end{align*}