Suppose that every ideal of $R$ is finitely generated, we want to show that $R$ is Noetherian.
Suppose that we have an ascending chain of ideals $$ \mathcal{A}_1 \subseteq \mathcal{A}_2 \subseteq \mathcal{A}_3\subseteq \dots $$ in $R,$Then $A = \bigcup_{i = 1}^{\infty} A_i$ is an ideal in $R$. Also, by our first assumption $A$ is finitely generated, say $A = \langle a_1, \dots , a_n \rangle.$
Now, assuming $a_{j} \in A_{i_{j}}$ for some $A_{i_{j}}$ where $j= 1,2, \dots ,n.$ Let $k = \operatorname{max}(i_1, \dots , i_n).$ Then $$a_1,a_2, \dots ,a_n \in A_{k}$$ because we have an ascending chain of ideals by assumption. And this implies that $$A = \langle a_1, \dots , a_n \rangle \subseteq A_k \subseteq A $$ Then we must have that $A = A_k.$
Thus, for any positive integer $i,$ we have that $$A_{k + i} \subseteq A = A_k \subseteq A_{k +i},$$
Hence, $ A_k = A_{k +i}$ for every positive integer $i,$ i.e. our ascending chain of ideals stabilizes. So, $R$ satisfies the $ACC$ condition and hence it is Noetherian as required.
Is my proof correct?