We know that a commutative ring $R$ is Noetherian, if for the increasing sequence of ideals $I_1\subseteq I_2 \subseteq…,\ \exists m\in \mathbb{Z}^+,\forall k\in \mathbb{Z}_{\geq m}:\ I_m=I_k$.
Question. If $R$ is Noehterian and every non-empty subset of ideals of $R$ has maximal element, then every ideal of $R$ is finitely
generated.
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?
Best Answer
The argument you are giving in no way requires $R$ to be unital. Note that if $R$ is not unital, then $\langle a\rangle$ still refers to the ideal generated by $a$. That means it is the smallest ideal which contains $a$, which is not necessarily the same as the set $Ra$. So, by definition, $a\in\langle a\rangle.$
(Note also that your statement of what you are proving seems to be mangled. You have never used the assumption that $R$ is Noetherian, nor do you need to. The assumption that every nonempty set of ideals has a maximal element is actually equivalent to assuming that $R$ is Noetherian.)