This is a mess; what you quote (up to but not including the last paragraph) is a perfectly fine proof that $R$ has maximal ideals; and it can easily be modified to prove the statement given. But as written, it is not a proof of the proposition that every nonempty collection of ideals of $R$ has a maximal element.
And the last part of the last paragraph is completely garbled. "The maximal ideal containing $I$" is problematic in several ways, not least the implicit assumption that there is a unique maximal ideal that contains $I$.
Of your first two questions, the first is "because they are doing things incorrectly" (or "because they are trying to prove that $R$ has maximal ideals).
For the second: since $a\in I_n$ and $I_n$ is an ideal, it contains $(a)$. But $(a)=I$. So we have
$$I = (a) \subseteq I_n \subseteq I_{n+1}\subseteq I,$$
and this shows that we have equality throughout. In particular, $I_n=I_{n+1}$.
Now, how to fix this argument to prove the desired result? Let $S$ be a nonempty collection of ideals of $I$. We want to show that $S$, ordered by inclusion, satisfies the hypotheses of Zorn's Lemma: every chain in $S$ has an upper bound in $S$. Let $\{I_j\}_{j\in J}$ be a chain in $S$. If $J=\varnothing$, then any element of $S$ (which exist, because $S$ is nonempty) is an upper bound.
(What follows is essentially their argument)
If $J$ is nonempty, let $I = \cup_{j\in J}I_j$. Then $I$ is an ideal of $R$, hence principal, $I=(a)$. Since $a\in\cup_{j\in J}$, there exists $n\in J$ such that $a\in I_n$. I claim that $I_n=I$ is an upper bound for $\{I_j\}_{j\in J}$ in $S$. Note that because $a\in I_n$ and $I_n$ is an ideal, then $(a)\subseteq I_n$. But $I_n\subseteq I=(a)$, so $I=I_n$.
Indeed, take any $j\in S$. We need to prove that $I_j\subseteq I_n$.
Since we have a chain, either $I_j\subseteq I_n$, or $I_n\subseteq I_j$. If $I_j\subseteq I_n$, there is nothing to do. If $I_n\subseteq I_j$, then
$$I = (a) = I_n \subseteq I_j\subseteq I = I_n,$$
so in fact $I_j=I_n$. Thus, $I_n$ is an upper bound for the chain. Finally, because $I_n\in S$, it is an upper bound in $S$.
This completes the verification that $S$ satisfies the hypotheses of Zorn's Lemma. Thus, $S$ has maximal elements, as desired.
The proof easily generalizes to rings in which every ideal is finitely generated. These are called Noetherian rings. In fact, the following are equivalent:
- Every ideal of $R$ is finitely generated.
- (ACC on ideals) Every ascending chain of ideals of $R$, $$I_1\subseteq I_2\subseteq\cdots\subseteq I_n\subseteq\cdots$$
stabilizes: there exists $n$ such that for all $k\geq n$, $I_n=I_k$.
- Every nonempty collection of ideals of $R$ has maximal elements.
Best Answer
Assuming strict descent (as otherwise any constant sequence is a counter-example). Let $a$ be the generator of the intersection. Then, $(a) \subset (a_i) \, \forall i$ and hence $a_i | a \, \forall i$. But since the descent is strict, none of the $a_i$'s are associates. Also $a_i | a_{i+1}$ so $\exists p_i$, a prime that divides $a_{i+1}$ but not $a_i$ for each $i$. That gives us infinitely many prime divisors for $a$, forcing $a=0$.