A Noetherian space is a finite union of irreducible closed subsets

Let $\mathcal{A}$ be the family of closed subsets of $X$ which are not a finite union of irreducible closed subsets of $X$, and suppose $\mathcal{A}\mathrm{\ne }\mathrm{\varnothing }$. I claim $\mathcal{A}$ has a minimal element [F5]. Otherwise every $F\in \mathcal{A}$ would have some $F^{\mathrm{\prime }}\in \mathcal{A}$ with $F^{\mathrm{\prime }}⊊F$, so the relation $R$ on $\mathcal{A}$ with $FR{F}^{\mathrm{\prime }}$ exactly when $F^{\mathrm{\prime }}⊊F$ would be entire on the nonempty set $\mathcal{A}$; choosing $F_0\in \mathcal{A}$ and applying dependent choice [F6], which is available by [F7] from the Axiom of Choice [F8] assumed in the statement, produces a sequence $F_0⊋F_1⊋F_2⊋\cdots$ of closed subsets of $X$, contradicting [F1]. Hence $\mathcal{A}$ has a minimal element $F$.

Let $F$ be a minimal element of $\mathcal{A}$ as in [step 1.1]. Then $F\mathrm{\ne }\mathrm{\varnothing }$, because the empty family is a finite union of irreducible closed subsets of $X$ and so $\mathrm{\varnothing }\mathrm{\notin }\mathcal{A}$; and $F$ is not irreducible, because an irreducible closed subset is the one-member union of itself. Since $F\mathrm{\ne }\mathrm{\varnothing }$, the failure of irreducibility [F2] supplies closed subsets $F_1,F_2$ of the subspace $F$ with $F=F_1\cup F_2$ and $F_1\mathrm{\ne }F\mathrm{\ne }{F}_2$; by [F3] there are closed subsets $G_1,G_2\subseteq X$ with $F_i=F\cap G_i$, so that $F_1$ and $F_2$ are closed in $X$ [F4], and $F_1⊊F$ and $F_2⊊F$.

By minimality of $F$ in $\mathcal{A}$, the strictly smaller closed subsets $F_1,F_2$ of [step 2.1] are not members of $\mathcal{A}$, so each of them is a finite union of irreducible closed subsets of $X$; the union of those two finite unions exhibits $F=F_1\cup F_2$ [F4] as a finite union of irreducible closed subsets of $X$, contradicting $F\in \mathcal{A}$. Hence $\mathcal{A}=\mathrm{\varnothing }$: every closed subset of $X$, and in particular $X$ itself, is a finite union of irreducible closed subsets of $X$.

Write $X=Y_1\cup \cdots \cup Y_m$ with each $Y_i$ irreducible and closed in $X$, as [step 3.1] provides. If some member is contained in the union of the others, delete the one of least index; each deletion lowers the number of members by one, so after finitely many deletions no choice is needed and one arrives at a subfamily $X=X_1\cup \cdots \cup X_n$ in which each $X_i$ is still irreducible and closed in $X$ and no $X_i$ is contained in ${\bigcup }_{j\mathrm{\ne }i}{X}_j$. By [F9] the irreducible components of $X$ are exactly $X_1,\dots ,X_n$; in particular $X$ has finitely many irreducible components, and the theorem together with the definition [F10] is proved. The Axiom of Choice is used at exactly one point, in [step 1.1], where it provides dependent choice [F7]; the splitting, the finite-union arguments and the deletion process use no choice principle. ∎