Axiom of choice in the proof that closed sets of Noetherian spaces are unions of finitely many irreducible subsets.

Hartshorne Proposition 1.5 states that any closed subset of a Noetherian topological space can be written as a union of finitely many closed irreducible subsets. A Noetherian topological space is a topological space that satisfies the descending chain condition on its closed subsets. I have one question regarding the proof of this theorem, and another question about an alternative proof.

The proof considers the set $\mathcal{S}$ of closed sets that cannot be written as a finite union of closed irreducible sets and shows $\mathcal{S}$ being nonempty results in a contradiction. It claims that if $\mathcal{S}$ is nonempty, then it has a minimal element. My first question is about this claim. Does this claim require the axiom of choice? I think it requires AOC, but it is not clear to me how.

My second question is about the following proof. Assume for contradiction that there exists a closed set $Y$ that cannot be written as a finite union of closed irreducible sets. So, $$Y=Y_1 \cup Y_2 \cup Y_3 \cup \cdots,$$ for some closed irreducible $Y_i$. Note that any set can be written as a union of singletons, which are closed and irreducible. I want to show the infiniteness of the union gives a contradiction. I was thinking that I am able to construct the chain $$Y \supset Y-Y_1 \supset Y-(Y_1\cup Y_2) \supset \cdots,$$ but this chain does not consist of closed sets. In contrast, the chain $$Y \supset Y_2\cup Y_3\cup Y_4 \cup \cdots \supset Y_3 \cup Y_4 \cup \cdots \supset \cdots,$$ consists of closed subsets, whereas I think I am not able to construct such a chain. The reason for this thought is that, in contrast to the first chain that at each step we require finitely many elements in hand, the second chain requires infinitely many elements at each step. I mean, for example, in the second step of the first chain we only require $Y$ and $Y_1$, while in the second step of the second chain, we require all $Y_i$; $i \in [2,\infty)$. Now, I ask myself why I think this may cause a problem when I specify the existence of $Y_i$ whose union gives $Y$. I am not able to answer myself; I just have a feeling that the construction of the second chain is not rigorous at this stage, and I may need to use the axiom of choice to make the existence of the second chain valid. Would you tell me whether I am thinking correctly? And how can we use the axiom of choice to show the existence of the second chain?