If C is a chain of non-principal ideals and the union of the ideals of the chain contains a generator why is the union then principal. I understand this is a contradiction already and seems painfully obvious but its not clicking for me at the moment.
Say S is a set of non-principal ideals and $\{C_i\}_{i \in \mathbb{N}}$ is a chain in S. and say theres a generator $d$ in some ideal $C_k$. Why is $\bigcup_{i \in \mathbb{N}} C_i = \langle d \rangle$ ?
Outwardly, one might guess you are working with Zorn's lemma on the poset of non-principal ideals.
Suppose for a moment that $\bigcup_{i \in \mathbb{N}} C_i$ is not in the poset: then it must be principal. But if $\bigcup_{i \in \mathbb{N}} C_i = ( d )$, then $d\in C_i$ for some $i$, and then $C_i\subseteq (d)\subseteq C_i$, so that we have equality and that $C_i$ is principal, a contradiction.
The upshot of this is that the union cannot be principal, and so the chain has an upper bound in the poset. That is, Zorn's lemma applies to the poset.
You appear to be transposing the assumption (that the union is principal) and the conclusion (that $d$ lies in one of the links in the chain) of this argument. This is possibly why you feel confused.