= Solution
An <Artinian module> is an $A$-module satisfying the <descending chain condition> on <submodules>: every chain $M_0\supseteq M_1\supseteq\cdots$ is eventually constant. An <Artinian ring> is a <ring> that is <Artinian> as a <module> over itself, so its descending chains of <ideals> stabilize.
We first prove the closure properties needed for finite generation. A <submodule> of an <Artinian module> is <Artinian>, because any descending chain in it is also a chain in the original <module>. A <quotient module> is <Artinian>, because inverse images turn a descending chain in the quotient into one upstairs. Conversely, consider a <short exact sequence>
$$
0\longrightarrow N\longrightarrow M\xrightarrow{\pi}Q\longrightarrow0
$$
with $N$ and $Q$ <Artinian>. For a descending chain $(M_i)$, the chains $M_i\cap N$ and $\pi(M_i)$ both stabilize. Choose $r$ beyond both stabilization indices. If $i\geq r$ and $x\in M_i$, equality of the images gives $y\in M_{i+1}$ with $\pi(y)=\pi(x)$. Then
$$
x-y\in M_i\cap N=M_{i+1}\cap N,
$$
so $x\in M_{i+1}$. Hence $M_i=M_{i+1}$. This proves the extension property of <Artinian modules in a short exact sequence>.
Apply this property inductively to $0\to A^{n-1}\to A^n\to A\to0$. Every finite <direct sum> $A^n$ is <Artinian> when $A$ is an <Artinian ring>. If $m_1,\ldots,m_n$ generate $M$, the map $(a_i)\mapsto\sum a_im_i$ is a <surjection> $A^n\twoheadrightarrow M$. The quotient property therefore proves
$$
\boxed{A\text{ Artinian and }M\text{ finitely generated}\ \Longrightarrow\ M\text{ Artinian}.}
$$
This argument also covers the zero <module>; it does not require a structure theorem for <Artinian rings>.
For a counterexample without finite generation, take any <field> $k$ and the <module>
$$
M=\bigoplus_{j\geq0}ke_j,\qquad M_i=\bigoplus_{j\geq i}ke_j.
$$
The <ring> $k$ is <Artinian>, since its only <ideals> are $0$ and $k$. But $M_i\supsetneq M_{i+1}$ for every $i$, since $e_i\notin M_{i+1}$. Thus \b[an infinite-dimensional <vector space> over a <field> provides the requested non-Artinian <module>].
Finally, let $f:M\to M$ be an injective <endomorphism> with $M$ <Artinian>. Its image chain stabilizes, so $f^rM=f^{r+1}M$ for some $r$. Given $x\in M$, choose $y$ with $f^rx=f^{r+1}y=f^r(fy)$. Since $f^r$ is injective, $x=fy$. Therefore
$$
\boxed{f\text{ injective}\ \Longrightarrow\ f\text{ surjective}.}
$$
This is precisely the assertion that <Artinian modules are co-Hopfian>.
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