I am currently going over the proof of the following proposition: (I know there is more than one post about this same result, but none can answer my current doubt).
Propostion. Given a commutative ring with identity $A$, the nilradical of $A$ is the intersection of all the prime ideals of $A$.
Proof. The inclusion $\subseteq $ is trivial to see. Conversely, let $f$ be a non-nilpotent element of $A$ and define the set
$$ S := \{a \text{ ideal of } A \colon f^n \notin a, \forall n \in \mathbb N \}.$$
$\color{red}{ \text{It is possible to show that } S = (S,\leqslant) \text{ has a maximal element (Zorn Lemma), where } \leqslant \text{ denotes the usual inclusion partial order relation.}}$
The proof goes on but the red text in my main doubt.
My attempt. I wish to show that, in fact, $S = (S,\leqslant) $ has a maximal element, by application of the Zorn Lemma. It is clear that $S \neq \emptyset$ since the null ideal $0$ is in $S$ (in fact, the null ideal only contains $0$ and $f^n \neq 0$, for every $n \in \mathbb N$ since $f$ is non-nilpotent). Therefore, let us consider an arbitrary chain $C$ in $S$ $(C \subseteq S)$ and consider the set $$C^\ast = \bigcup_{I \in C} I.$$ It is easy to see that $C^\ast $ verifies that $f^n \notin C^\ast,$ since $f^n \notin I,$ for every $I \in C$ and by definition of union of sets. $\color{red}{ \text{Altought, I don't know how to show that $C^\ast$ is, in fact, an ideal (the union of ideals is not necessarily an ideal)}}$.
Assuming $C^\ast$ is an ideal, we just guaranteed that $C^\ast \in S$ and it is of trivial observation that $I \subseteq C^\ast$ (again, just by definition of union of sets). Therefore, $I \leqslant C^\ast,$ for every $I \in C$ and so $C^\ast$ is an upper bound for the arbitrary chain $C$. Hence, every chain has an upper bound and by Zorn Lemma we can guarantee that $S$ has a maximal element.
Although the union of ideals is in general not an ideal, the union of a chain of ideals is always an ideal.
Let $C$ a chain of ideals and $C^*$ the union of the elements of $C$. We can easily check that $C^*$ satisfies the closure properties of an ideal.
Scalar closure: Let $a \in A$ and $c \in C^*$. Then there exists an ideal $I \in C$ such that $c \in I$. Since $I$ is an ideal, $ca \in I \subseteq C^*$. Note that we didn't need $C$ to be a chain for this part.
Additive closure: If $c_1, c_2 \in C^*$ then we can find $I_1, I_2$ in $C$ such that $c_1 \in I_1, c_2 \in I_2$. Because $C$ is a chain, it is totally ordered, and therefore either $I_1 \subseteq I_2$ or $I_2 \subseteq I_1$. Without loss of generality, say $I_1 \subseteq I_2$. Then $c_1, c_2 \in I_2$, and since $I_2$ is an ideal, also $c_1 + c_2 \in I_2 \subseteq C^*$.