An answer turns up as a byproduct in this solution written for the Noetherian version of this question. But we can probably argue more directly given the Artinian condition too.
Suppose $M$ is not a finite direct sum of indecomposable modules. Then $M$ itself is not indecomposable, yielding a proper decomposition $A\oplus B$. Clearly it can't be the case that both $A$ and $B$ decompose into finitely many indecomposable submodules, because taking them all together would make a finite decomposition for $M$. Let $C$ be one that doesn't decompose that way.
Notice what we have observed is that a module that doesn't decompose into finitely many indecomposables always has a proper submodule with the same property. Using this observation, we inductively define a properly descending chain $C_1\supset C_2\supset C_3\supset \cdots$, which obviously can't happen in an Artinian module. The module would have to be non-Artinian. This proves the contrapositive of the original statement to be proven.