Let $M$ be any $R$-module. Let $A$ be the submodule generated by all artinian submodules of $M$. Show any nonzero submodule of $A$ contains at least one simple submodule.
My attempt: By definition, $A=\sum_{S \leq M} S$, where $S$ is artinian. But this is also $A=\bigcap_{K \subseteq T,K \text{ artinian}}T$; that is, it is the intersection of all submodules of $M$ containing every artinian submodule of $M$. So if $S \leq A$, then it is a submodule of every submodule of $M$ containing all artinian submodules. So in particular, it contains an artinian submodule. This submodule contains a simple submodule, which is then a submodule of $S$.
Is this really that simple?
Your statement about the intersection is quite dubious and the argument is unfortunately wrong.
Prove that, if $L$ is a nonzero submodule of $A$ and $0\ne x\in L$, then $xR$ (or $Rx$ if you're dealing with left modules) is artinian.
Every nonzero artinian module has a simple submodule.