[proofplan]
We prove the stronger existence statement that every nonempty closed subset of $X$ is a finite union of nonempty irreducible closed subsets. If a counterexample existed, Noetherianity would provide a minimal one; that minimal counterexample cannot be irreducible, so it splits into two proper closed subsets, both decomposable by minimality. For uniqueness, we compare two irredundant decompositions and use irreducibility to show each component in one decomposition is contained in a component of the other, then use irredundancy to force equality.
[/proofplan]
[step:Reduce existence to a minimal closed counterexample]
We use the convention that an irreducible [topological space](/page/Topological%20Space) is nonempty. Let $\mathcal{C}$ be the collection of nonempty closed subsets $Z\subset X$ which cannot be written as a finite union of nonempty irreducible closed subsets of $X$.
Suppose, for contradiction, that $\mathcal{C}$ is nonempty. Since $X$ is Noetherian, every nonempty family of closed subsets of $X$ has a minimal member under inclusion: otherwise one could choose a strictly descending sequence of closed subsets, contradicting the descending chain condition. Hence there exists $Z\in\mathcal{C}$ such that no proper nonempty closed subset of $Z$ lies in $\mathcal{C}$.
[guided]
We want to prove that every nonempty closed subset of $X$ has the desired finite decomposition. The standard Noetherian strategy is to argue by minimal counterexample.
Define $\mathcal{C}$ to be the set of all nonempty closed subsets $Z\subset X$ for which no finite decomposition
\begin{align*}
Z=Z_1\cup\cdots\cup Z_m
\end{align*}
exists with each $Z_\ell$ a nonempty irreducible closed subset of $X$. We assume $\mathcal{C}$ is nonempty and seek a contradiction.
The Noetherian hypothesis is used here in its closed-set form: descending chains of closed subsets stabilize. This implies that every nonempty family of closed subsets has a minimal member under inclusion. Indeed, if a nonempty family had no minimal member, then after choosing one member $F_1$, non-minimality would give $F_2$ in the family with $F_2\subsetneq F_1$, then $F_3\subsetneq F_2$, and so on. This produces a strictly descending chain of closed subsets, contradicting Noetherianity.
Applying this to $\mathcal{C}$ gives a nonempty closed subset $Z\subset X$ which is a minimal counterexample: $Z$ itself has no finite irreducible closed decomposition, while every proper nonempty closed subset of $Z$ does have one.
[/guided]
[/step]
[step:Split the minimal counterexample into decomposable proper closed subsets]
The minimal counterexample $Z$ is not irreducible, because if it were irreducible then $Z$ itself would be a finite union of one nonempty irreducible closed subset, contradicting $Z\in\mathcal{C}$.
Thus $Z$ is reducible. Therefore there exist proper closed subsets $A,B\subsetneq Z$, closed in the [subspace topology](/page/Subspace%20Topology) on $Z$, such that
\begin{align*}
Z=A\cup B.
\end{align*}
Since $Z$ is closed in $X$, the subsets $A$ and $B$ are closed in $X$. If either $A$ or $B$ is empty, omit it. Every remaining subset is a proper nonempty closed subset of $Z$, so by minimality of $Z$ it is a finite union of nonempty irreducible closed subsets of $X$. Combining these finite unions gives a finite irreducible closed decomposition of $Z$, contradicting $Z\in\mathcal{C}$.
Therefore $\mathcal{C}$ is empty. Since $X$ itself is nonempty and closed in $X$, there exist nonempty irreducible closed subsets $X_1,\dots,X_r\subset X$ such that
\begin{align*}
X=X_1\cup\cdots\cup X_r.
\end{align*}
[/step]
[step:Show an irreducible set in a finite closed union lies in one member]
Let $W\subset X$ be a nonempty irreducible subset with the subspace topology, and let $F_1,\dots,F_m\subset X$ be closed subsets such that
\begin{align*}
W\subset F_1\cup\cdots\cup F_m.
\end{align*}
Then
\begin{align*}
W=(W\cap F_1)\cup\cdots\cup(W\cap F_m),
\end{align*}
where each $W\cap F_j$ is closed in $W$. We claim that $W\subset F_j$ for some $j$.
For $m=1$ this is immediate. For $m=2$, this is exactly irreducibility: $W$ cannot be the union of two proper closed subsets. For $m>2$, write
\begin{align*}
W=(W\cap F_1)\cup\left(W\cap(F_2\cup\cdots\cup F_m)\right).
\end{align*}
Both displayed subsets are closed in $W$. By irreducibility, either $W\subset F_1$ or
\begin{align*}
W\subset F_2\cup\cdots\cup F_m.
\end{align*}
In the second case, induction on $m$ gives $W\subset F_j$ for some $j\in\{2,\dots,m\}$. This proves the claim.
[guided]
The uniqueness argument will repeatedly use the following finite-union consequence of irreducibility. Suppose a nonempty irreducible subset $W\subset X$ is covered by finitely many closed subsets $F_1,\dots,F_m$ of $X$:
\begin{align*}
W\subset F_1\cup\cdots\cup F_m.
\end{align*}
Intersecting the cover with $W$ gives
\begin{align*}
W=(W\cap F_1)\cup\cdots\cup(W\cap F_m).
\end{align*}
Each set $W\cap F_j$ is closed in $W$ because $F_j$ is closed in $X$ and $W$ has the subspace topology.
The definition of irreducibility forbids writing $W$ as the union of two proper closed subsets. To pass from two closed subsets to finitely many, we use induction. If $m=1$, then $W\subset F_1$. If $m=2$, irreducibility says that one of $W\cap F_1$ and $W\cap F_2$ must equal all of $W$, hence $W\subset F_1$ or $W\subset F_2$.
For $m>2$, group the last $m-1$ closed subsets together:
\begin{align*}
W=(W\cap F_1)\cup\left(W\cap(F_2\cup\cdots\cup F_m)\right).
\end{align*}
The set $F_2\cup\cdots\cup F_m$ is closed in $X$ because it is a finite union of closed subsets, so both terms in this union are closed in $W$. Irreducibility gives either $W=W\cap F_1$, which means $W\subset F_1$, or
\begin{align*}
W=W\cap(F_2\cup\cdots\cup F_m),
\end{align*}
which means $W\subset F_2\cup\cdots\cup F_m$. In the second case, the induction hypothesis applied to $F_2,\dots,F_m$ gives $W\subset F_j$ for some $j\in\{2,\dots,m\}$. Thus in every case one closed member contains $W$.
[/guided]
[/step]
[step:Compare two irredundant decompositions component by component]
Let
\begin{align*}
X=X_1\cup\cdots\cup X_r=Y_1\cup\cdots\cup Y_s
\end{align*}
be two irredundant decompositions into nonempty irreducible closed subsets.
Fix $i\in\{1,\dots,r\}$. Since $X_i\subset X=Y_1\cup\cdots\cup Y_s$, the previous step applied to $W=X_i$ and $F_j=Y_j$ gives an index $j\in\{1,\dots,s\}$ such that
\begin{align*}
X_i\subset Y_j.
\end{align*}
Now apply the same argument to $W=Y_j$ and the closed cover $X=X_1\cup\cdots\cup X_r$. There exists $k\in\{1,\dots,r\}$ such that
\begin{align*}
Y_j\subset X_k.
\end{align*}
Hence
\begin{align*}
X_i\subset Y_j\subset X_k.
\end{align*}
By irredundancy of the decomposition $X=X_1\cup\cdots\cup X_r$, the containment $X_i\subset X_k$ forces $k=i$. Therefore
\begin{align*}
X_i=Y_j.
\end{align*}
Thus every $X_i$ occurs among the $Y_j$. Reversing the roles of the two decompositions shows that every $Y_j$ occurs among the $X_i$. Since neither decomposition contains repeated components, again by irredundancy, the two finite lists have the same length and agree up to reordering. Hence $r=s$ and, after reindexing, $X_i=Y_i$ for every $i\in\{1,\dots,r\}$. This completes the proof.
[/step]
