Let $N$ be an $A$-submodule of $M$. It suffices to prove that $N$ is finitely generated. We prove this by induction on the number of generators of $M$. Suppose $M = Ax_1 + \cdots + Ax_n$. If $n = 1$, $M$ is isomorphic to $A/I$, where $I$ is a left ideal of $A$. Hence $N$ is finitely generated. Suppose $n > 1$. Let $L = Ax_1 + \cdots + Ax_{n-1}$. There exists the following exact sequence
$0 \rightarrow N \cap L \rightarrow N \rightarrow M/L.$
By the induction hypothesis, $N \cap L$ is finitely generated. Since $M/L$ is generated by the image of $x_n$, the image of $N \rightarrow M/L$ is finitely generated. Hence $N$ is finitely generated by the following lemma.
Lemma Let $A$ be a ring. Let $M$ be an $A$-module. Let $N$ be an $A$-submodule of $M$. Suppose $N$ and $M/N$ are finitely generated. Then $M$ is also finitely generated.
Proof: Suppose $N$ is generated by $x_1,\dots,x_n$ and $M/N$ is generated by $y_1$ mod $N,\dots,y_m$ mod $N$. Let $x \in M$. Then there exist $b_1,\dots,b_m \in A$ such that $x \equiv b_1y_1 + \cdots + b_my_m$ (mod $N$). Hence there exist $a_1,\dots,a_n \in A$ such that $x - (b_1y_1 + \cdots + b_my_m) = a_1x_1 + \cdots + a_nx_n$. Hence $M$ is generated by $x_1,\dots,x_n, y_1,\dots,y_m$. QED