A Noetherian space is a disjoint union of finitely many connected closed subsets (i.e., its connected components). This is the first part of Exercise 1.2.8(1a) of Springer's, "Linear Algebraic Groups (Second Edition)".
The Details:
Since definitions vary:

A topological space $(X,\tau)$ is a set $\tau$ of subsets of $X$, called closed subsets, such that

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*$\varnothing, X\in\tau$,

*The intersection $$\bigcap_{i\in I}X_i$$ of any closed subsets $(X_i)_{i\in I}$ is closed, where $I$ is arbitrary, and

*The union of finitely many closed sets is closed.


Note that $\tau$ is omitted sometimes when the context is clear.
On page 2 of Springer's book, paraphrased, we have this:

A topological space $X$ is Noetherian if any family of closed subsets of $X$ contains a minimal one.

On page 4, we have:

[A] topological space is connected if it is not the union of two disjoint proper closed subsets.

The Question:

Show that each noetherian space $X$ is a disjoint union of finitely many connected closed subsets.

Context:
I have no clue how to solve this.
To provide context, I will answer the questions given here:

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*What are you studying?

A postgraduate research degree in linear algebraic groups.

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*What text is this drawn from, if any? If not, how did the question arise?

(See above.)

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*What kind of approaches (to similar problems) are you familiar with?

I have very little experience with this sort of thing. Here is a recent question of mine:
The components of a Noetherian space are its maximal irreducible closed subsets.

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*What kind of answer are you looking for? Basic approach, hint, explanation, something else?

I would very much appreciate a full answer, although I am happy to have some strong hints.

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*Is this question something you think you should be able to answer? Why or why not?

No. Topology is not my forté.
Please help :)
 A: As noted in the linked question, a Noetherian space $X$ has finitely many maximal irreducible components and the union of these components covers $X$. By definition of irreducibility, each of these irreducible components must be connected. Let $C_1, \dots, C_n$ be the maximal irreducible components of $X$. Form the set $U_1$ as follows: Take $C_1$ and if there is any $C_{i_1}$ with $C_{i_1} \cap C_1 \neq \varnothing$, take $U_1$ to be $C_1 \cup C_{i_1}$. If there is any remaining $C_{i_2}$ with $C_{i_2} \cap U_1 \neq \varnothing$, redefine $U_1$ as $U_1 \cup C_{i_2}$. Since there are finitely many maximal irreducible components of $X$, this process will terminate. Furthermore, $U_1$ will be closed and be connected. (It will be the finite union of closed sets, and if it were not connected then it could be described as the disjoint union of two nonempty closed sets, which would imply that at least one of the $C_i$'s in the union defining $U_1$ was the union of two closed sets, contradicting its irreducibility). If necessary, repeat this process. Since there were finitely many $C_i$'s to begin with, this process will terminate. The resulting $U_i$'s will all be closed and connected.
