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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?

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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.

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    Yes, it felt far to swift to me. Using your idea, would it be just the fact that $x=s_1+\cdots+s_n$, where $s_i \in S_i$ is artinian. Then I have the exact sequence: $0 \to Rx \to S_1+\cdots+S_n \to (\sum S_i)/(Rx)$. Then $Rx$ is artinian being the submodule of an artinian module? – 2017-01-09
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    @NPH Yes: $x$ belongs to a finite sum of artinian submodules of $M$ and a sum of a finite number of artinian submodules is artinian. – 2017-01-09
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    I didn't try your line of thought because trying to show a submodule of a submodule of a module generated by artinian submodules was artinian started to kill my head and seemed like a train of thought not worth pursuing. But taking your suggestion, took me all of 30 seconds! How wrong I was! Thanks! – 2017-01-09