This is what I want to prove:
Every family $\mathcal{C}$ of ideals in $R$ has a maximal element in $\mathcal{C}$ implies that every ideal in $R$ is finitely generated.
I found this answer here Every subset of ideals of $R$ has maximal element $\implies$ Every ideal of $R$ is finitely generated, if $R$ is Noetherian.
Proof. We take an ideal $I$ of $R$ and we define the set $$F:=\{J: J\text{ finitely generated ideal of } R \text{ and } J\subseteq I\}.$$ We can see that $F\neq \emptyset$ (because $\langle 0_R \rangle \in F$). So, from hypothesis, there exists a maximal element $M\in F$.
We want to show that $I=M$.
Obviously, $M\subseteq I$, because $M\in F$. We suppose now that $M\subsetneq I$. Then, there exists an element $a\in I\backslash M\iff a\in I$, with $a\notin M$.
We consider the ideal $$J:=M+\langle a\rangle.$$ Then, if $M=\langle m_1,...,m_n \rangle$, we have $J=M+\langle a\rangle=\langle m_1,..., m_n,a \rangle \implies J\in F$
Question: Why $M\neq M+\langle a \rangle $?
If $M= M+\langle a \rangle$ $\underline{ \text{and}}$ $R$ has $1_R$, then $a=1_R\cdot a \in M+\langle a \rangle =M $, contradiction.
But what happens if $R$ hasn't unity?
Here is my correction to this answer:
Assume that every family $\mathcal{C}$ of ideals in $R$ has a maximal element in $\mathcal{C}$ relative to the partial order of set inclusion. Let $\mathcal{A}$ be any ideal of $R$.Define $$F:=\{I: I\text{ finitely generated ideal of } R \text{ and } I\subseteq \mathcal{A}\}.$$
Now, It is clear from the definition of $F$ that $F\neq \emptyset$ (because $\langle 0_R \rangle \in F$). Now, by our assumption, there exists a maximal element $M\in F$ relative to the partial order by set inclusion.
We want to show that $\mathcal{A} = M$. If $M\neq \mathcal{A}$, then there is some $\langle a \rangle \in A\setminus M$, and then $M\cup \langle a \rangle$ is a finite subset of $A$ which is a proper superset of $M$. But $M$ was maximal, so no such $\langle a \rangle$ can exist, so $\mathcal{A}=M$.
Still I have a wrong step in my correction, which is considering this $M\cup \langle a \rangle$ as an ideal, I know that the union of 2 ideals is not necessarily an ideal unless one of them is contained into the other, I feel like I should consider the ideal $J:=M+\langle a\rangle $ instead, but then how can I show that this ideal is generated by $\langle m_1,..., m_n,a \rangle$? How can I show that it contains the ideal $M$ and not equal to it i.e. $M \subsetneq J$? that could anyone help me correct that step in my proof please?
From what I can discern from your comments, the following question remains unresolved: given a commutative ring $R$ (possibly without unity $1_R$), an ideal $M \subseteq R,$ and an element $a \in R - M,$ why is it true that $M \subsetneq M + aR?$ On the contrary, let us assume that $M = M + aR.$ Consequently, for every element $ar$ in $aR,$ there exist elements $m, n \in M$ such that $m = n + ar$ so that $ar = m - n.$ By hypothesis that $M$ is an ideal, it follows that $ar = m - n \in M$ so that $aR \subseteq M.$ Obviously, this is a contradiction if $R$ is unital: we would have that $a = a \cdot 1_R \in M.$
Unfortunately, there is no problem here if the ring is non-unital. For instance, it is trivial to see that $4 \mathbb Z = 4 \mathbb Z + 4 \mathbb Z = 4 \mathbb Z + 2(2 \mathbb Z)$ for $R = 2 \mathbb Z,$ $M = 4 \mathbb Z,$ and $a = 2.$ Explicitly, we have that $4i = 4 \cdot 0 + 2(2i)$ for all integers $i$ so that $4 \mathbb Z \subseteq 4 \mathbb Z + 2(2 \mathbb Z)$ and $4i + 2(2j) = 4(i + j)$ for all integers $i$ and $j$ so that $4 \mathbb Z + 2(2 \mathbb Z) \subseteq 4 \mathbb Z.$ Of course, $2 \mathbb Z$ is a non-unital ring; $4 \mathbb Z$ is an ideal in $2 \mathbb Z$ because $4i - 4j = 4(i - j) \in 4 \mathbb Z$ and $(2i)(4j) = 4(2ij) \in 4 \mathbb Z$ for all integers $i$ and $j$; and $2$ is not in $4 \mathbb Z$ by definition.
Consequently, the assumption that $R$ is a unital ring is crucial to the proof that $M \subsetneq M + aR.$ But it is standard to assume that all rings are unital; a non-unital ring is more commonly referred to as a rng (no "$i$" means no unity).