Yes, that’s what I had in mind.

Um is what you said equivalent to if $\mathcal M$ is an irreducible component of $X$, then $\mathcal M = \bigcup (\mathcal M \cap X_i)$, hence $\mathcal M \cap X_i = \mathcal M$ for some $i$, so that $\mathcal M \subseteq X_i$ hence $\mathcal M = X_i$ by maximality? Because you seem to “see it” but I wonder if you had the same details in mind or something even more straight forward.

“It follows that Noetherian spaces have a unique decomposition into finitely many irreducible components. ”

Why? You’ve only proven that the space can be written as a union of finitely many irreducible components, you don’t know if there are finitely many irreducible components at all, so we don’t know about uniqueness either… maybe I’m missing something but I don’t see it.

“a) It suffices to show that the closure of an irreducible subspace is irreducible. Given S \subset X irreducible write \bar{S} = X_1 \cup X_2 where X_1, X_2 are closed in \bar{S}. Then the restrictions of X_1, X_2 to S are closed in the subspace topology, so S is reducible unless one of X_1, X_2 is disjoint from S. Then the other must contain S, and since it is closed it must be all of \bar{S}, so one of X_1, X_2 is empty. ”

feels a little wrong to me (I still agree with the statement though). I don’t understand how you get empty sets in there. My proof goes like this : if \bar{S} = X_1 \cup X_2, then S = (S \cap X_1) \cup (S \cup X_2) where (S \cap X_i) are closed in the subspace topology, hence S \cap X_1 = S or S \cap X_2 = S, i.e. S \subseteq X_1 or S \subseteq X_2. Since X_1 and X_2 are closed in \bar{S}, it means X_1 \cup X_2 = S \subseteq X_1 or X_1 \cup X_2 = S \subseteq X_2, i.e. S = X_1 or S = X_2. I don’t get any empty sets anywhere.

Can you either explain your proof more in detail or agree with my comment? I’m still a little confused with this irreducible component story…