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Let $M$ be a finite $R$-module that is noetherian and such that $M_{\mathfrak{p}}$ is artinian for each $\mathfrak{p}\in \text{Spec}(R)$. Then $M$ is an artinian $R$-module.

I have tried using a proof with Zorn's lemma: Let $\Gamma = \{ N: N\subseteq M \text{ is artinian } \}$. Since $0\in \Gamma$, and $M$ is noetherian, $\Gamma$ contains a maximal element, call it $N$. But then $N_{\mathfrak{p}}$ is a maximal artinian submodule of $M_{\mathfrak{p}}$ (I don't think this is true (a priori); I think proving this is equivalent to proving the wanted statement) for all $\mathfrak{p}$, so $N_{\mathfrak{p}} = M_{\mathfrak{p}}$ for all $\mathfrak{p}$, implying $N=M$.

I have also tried showing Jac$(R)M = \mathfrak{m}_1\cap \dots \cap \mathfrak{m}_t M$ where $\mathfrak{m}_t$ are finitely many maximal ideals, with which I have had no luck.

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    Consider a chain of submodules in M. If you localize at each prime the resulting chain stabilizes. Show that finiteness allows you to show that *all* the chains corresponding to all primes stabilize at the same point. – 2017-02-01
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    First you can reduce to cyclic modules, then you can reduce to $M=R$, and then noetherianness lets you replace artinian by zero-dimensional. The dimension of a ring can of course be tested locally. – 2017-02-01

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The sum of two artinian submodules is artinian. Hence - by induction on the number of generators - you can assume that $M$ is cyclic, i.e. $M = R/\operatorname{Ann}(M)$. Of course $R/\operatorname{Ann}(M)$ is artinian as a $R$-module if and only if it is artinian as a $R/\operatorname{Ann}(M)$-module (i.e. as a ring).

So we have the following situation: We have a ring $A$, which is noetherian and locally artinian. We want to prove that it is artinian.

Because of the noetherianess of $A$ and all its localizations, we have artinian = zero-dimensional and then the claim follows from $$\dim A = \sup\limits_{\mathfrak p \in \operatorname{Spec} A} \dim A_{\mathfrak p}.$$