You state no question, so I am not sure what you are asking. The argument Eisenbud seemed to have in mind is flawed, as you noticed. However the result is true:

Let $I$ be the maximal ideal of $B$, and $k = B / \mathfrak{m}$. The $B[t_1,\dots,t_n]$-module $D^n(B)$ has a finite filtration by the submodules $I^r D^n(B)$, with successive quotients $D^n(k) \otimes_k I^r/I^{r+1}$ where $I^r/I^{r+1}$ is a finite dimensional $k$-vector space, so it is enough to show that $D^n(k)$ is an artinian $k[t_1,\dots,t_n]$-module. Any sub- $k[t_1,\dots,t_n]$-module of $D^n(k)$ is also a sub-$k[[t_1,\dots,t_n]]$-module, and the annihilator map \begin{align*} \{ k[t_1,\dots,t_n]-submodules \ of \ D^n(k) \} & \rightarrow \{ ideals \ of \ k[[t_1,\dots,t_n]] \} \end{align*} is injective and inclusion reversing. Since $k[[t_1,\dots,t_n]]$ is a noetherian ring, we get the conclusion.

You state no question, so I am not sure what you are asking. The argument Eisenbud seemed to have in mind is flawed, as you noticed. However the result is true:

Let $I$ be the maximal ideal of $B$, and $k = B / \mathfrak{m}$. The $B[t_1,\dots,t_n]$-module $D^n(B)$ has a finite filtration by the submodules $I^r D^n(B)$, with successive quotients $D^n(k) \otimes_k I^r/I^{r+1}$ where $I^r/I^{r+1}$ is a finite dimensional $k$-vector space, so it is enough to show that $D^n(k)$ is an artinian $k[t_1,\dots,t_n]$-module. Any sub- $k[t_1,\dots,t_n]$-module of $D^n(k)$ is also a sub-$k[[t_1,\dots,t_n]]$-module, and the annihilator map \begin{align*} \{ k[t_1,\dots,t_n]\text{-submodules of } D^n(k) \} & \rightarrow \{ \text{ideals of } k[[t_1,\dots,t_n]] \} \end{align*} is injective and inclusion reversing. Since $k[[t_1,\dots,t_n]]$ is a noetherian ring, we get the conclusion.