Let $R$ be a Dedekind ring, which means
- integral domain,
- integrally closed,
- Noetherian, which means that given any chain of ideals in $R$: $$I_1\subseteq\cdots \subseteq I_{k-1}\subseteq I_{k}\subseteq I_{k+1}\subseteq\cdots$$ there exists an $n$ such that: $$I_{n}=I_{n+1}=\cdots\,.$$
- and for which every prime ideal $\mathfrak{p} \neq 0$ is maximal.
Let $I\subset R$ be a prime/maximal ideal. In a proof suddenly the assertion is made that $I$ is finitely generated, e. g. $$I = \{r_1 e_1 + \cdots + r_n e_n : r_1,\ldots, r_n \in R \}\, .$$
Is that because otherwise we could construct an ever increasing chain of ideals $$I_1 \subsetneq I_2 \subsetneq I_3 \subsetneq \cdots $$ with $I_k = (e_1, \ldots, e_k)$, which would be a contradiction to $R$ being Noetherian?
Indeed, an alternative definition of Noetherian ring is a ring where every ideal is finitely generated.
As you noticed, a ring $R$ admits an infinitely generated ideal if and only if there is an infinite ascending chain of ideals.
Indeed, suppose there is an ideal of $R$ which can only be generated by an infinite set $\{e_i\}_{i \in I}$ and extract a sequence $(e_n)_{n \in \Bbb{N}}$ from this set. Then you have the chain $$ (e_1) \subsetneq (e_1,e_2) \subsetneq (e_1,e_2,e_3) \subsetneq \dotsb $$ On the other hand, if you have an infinite ascending chain $$ I_1 \subsetneq I_2 \subsetneq I_3 \subsetneq \dotsb $$ you can find a sequence of elements $\{e_k\}_{k \in \Bbb{N}}$ such that $e_1 \in I_1$ and $e_k \in I_k \setminus I_{k-1}$ for $k > 1$. Then you can check that the ideal $J = ( e_1,e_2,\dotsc )$ cannot be generated by a finite subset of $\{e_k\}_{k \in \Bbb{N}}$.