 Solution

Work with the usual commutative unital Noetherian ring convention. First prove that every submodule of $R^d$ is a finitely generated module. For $d=1$, a submodule is an ideal, hence finitely generated by Noetherianity. For the induction step, project a submodule $N\subseteq R^d$ onto its last coordinate. Its image is a finitely generated ideal; choose lifts in $N$ of its generators. The projection kernel lies in $R^{d-1}$ and is finitely generated by induction. The kernel generators together with the chosen lifts generate $N$, since subtracting a combination of the lifts leaves an element of the kernel.

Now choose a surjection $R^d\to M$ from a finite generating set of the module $M$. For any submodule $U\subseteq M$, its inverse image in $R^d$ is finitely generated by the result just proved. The images of those generators generate $U$.

Every submodule of $M$ is finitely generated. In particular $M$ is a Noetherian module: the union of an ascending chain is a submodule, and its finite generators all belong to a single member of the chain, forcing eventual stabilization.