Perhaps I should spell out the argument given in the linked answer, and you can let me know which step(s) you query.
(1) Let $M$ be as given in your question. If $M$ is Noetherian, then as $M$ is also faithful, it will give you an answer to your question. Suppose that $M$ is not Noetherian. We will deduce a contradiction.
(2) By the ACC condition given, we may pick a maximal submodule of the form $IM\subseteq M$ where $I\subset A$ an ideal, and $M/IM$ not Noetherian.
(3) Let $I_0={\rm Ann}(M/IM)$. Then $I\subseteq I_0$ and $I_0M\subseteq IM$ by definition of annihilator. Thus $IM=I_0M$. Let $M'=M/I_0M$. Then $M'$ is faithful as a module over $A'=A/I_0$.
(4) $M'$ is not Noetherian over $A$ by construction. It has the same submodules when regarded as a module over $A'$, so it is not Noetherian over $A'$ either.
(5) However for any ideal $0\neq J\subset A'$, we have $M'/JM'=M/KM$ for some ideal of $A$: $K\supsetneq I_0\supset I$. Here $K$ is the preimage of $J$ under the quotient map $A\to A/I_0$. By the maximality condition on $I$, we have that $M/KM$ is Noetherian over $A$, so $M'/JM'$ is Noetherian over $A'$.
(6) Suppose that $N\subseteq M'$ is a submodule (over $A'$). Either $M'/N$ is faithful, or it is not, in which case $N\supset iM'$ for some $0\neq i\in A'$, so $M'/N$ is a quotient of $M'/iM'$, hence Noetherian by (5).
(7) Given a nested chain of submodules $\{N_i\subseteq M'|i\in \mathcal{I}\}$, with $M'/N_i$ faithful over $A'$, let $N$ be the union of the $N_i$. If $aM'\subseteq N$, for some $0\neq a\in A'$, then for $s=1,2,\cdots,r$, we have $am_s\in N$, so we have $am_s\in N_{i_s}$ for some $i_s\in \mathcal{I}$. Let $i$ denote the supremum of $i_1,i_2,\cdots,i_r$. Then we have $aM\subseteq N_i$, contradicting the assumption that $M/N_i$ is faithful for all $i\in \mathcal{I}$. Thus $M'/N$ is faithful.
(8) By (3) we know that $M'/\{0\}$ is faithful. By (7) we know that modules $N$ with $M/N$ faithful are closed under nested unions. Thus by Zorn's lemma, we may choose $N$ maximal such that $M'/N$ is faithful.
(9) Let $M''=M'/N$. By the maximality of $N$, we know that any proper quotient of $M''$ is not faithful, hence by (6) such a quotient must be Noetherian.
(10) Let $$\{0\}\subsetneq N_1\subsetneq N_2\subsetneq N_3\subsetneq\cdots$$ be a strictly increasing infinite chain of submodules in $M''$. Then $$N_2/N_1\subsetneq N_3/N_1\subsetneq N_4/N_1\subsetneq\cdots$$ is a strictly increasing infinite chain of submodules of $M''/N_1$, contradicting (9).
(11) Now we know that $M''$ is both faithful and Noetherian (over $A'$), we will deduce that $A'$ is Noetherian: Suppose we have an infinite strictly increasing chain of ideals $I_i$ in $A'$. For fixed $s=1,2,3,\cdots,r$, we know that the submodules $I_im_s\subseteq M''$ will eventually be constant. Thus we have an infinite strictly ascending chain of ideals $I_i\cap {\rm Ann} (m_s)\subseteq {\rm Ann} (m_s)$ (Important: Here we regard $m_s\in M''$). Repeating this step for $s=1,2,3,\cdots,r$, we obtain an infinite strictly ascending chain of ideals inside $${\rm Ann} (m_1)\cap {\rm Ann} (m_2)\cap {\rm Ann} (m_3)\cap\cdots\cap {\rm Ann} (m_r).$$
However this intersection is $\{0\}$ as $M''$ is faithful over $A'$. This gives us a contradiction to the existence of the infinite strictly increasing chain of ideals in $A'$. We conclude $A'$ is Noetherian.
(12) As $A'$ is Noetherian, and $M'$ is finitely generated over $A'$, we have that $M'$ is Noetherian over $A'$, contradicting (4).
(13) This is the contradiction we were seeking in (1). Thus we know that $M$ is both faithful and Noetherian over $A$, answering your question.
Best Answer
The argument is a proof by contradiction. Here is a rough outline of the strategy: Suppose there is a counterexample $M$ over a ring $A$ (meaning that $M$ satisfies the hypotheses of the theorem but is non-Noetherian). Then we can construct from this one counterexample $M$ another counterexample $M'$ over another ring $A'$ that has an additional property (that $M'/IM'$ is Noetherian for any non-zero ideal $I$ of $A'$). The remainder of the proof then derives a contradiction using this additional property.
So, what this shows is that there's no counterexample that has this additional property. But the proof starts by showing: if there is any counterexample whatsoever to the theorem, then there is a counterexample with that additional property. So the theorem can't have any counterexample, which completes the proof by contradiction.
Of course, if this were a direct proof, then the strategy wouldn't make sense, because you'd only be proving something about modules of that particular form. But since it's a proof by contradiction, we just need to start by assuming there's a counterexample and derive a contradiction from this, and it's perfectly fine if one of the intermediate steps is "show that the existence of a counterexample implies the existence of another counterexample, and consider this other counterexample instead".