I'm not sure if the cited argument can be repaired. Nonetheless, the claim is true. First, note that $\mathbb{A}^n$ is a noetherian topological space, in the sense that every strictly descending chain of closed subspaces must be finite. This is a straightforward consequence of the fact that a polynomial ring over a field is noetherian as a ring.
Proposition. Let $X$ be a topological space. The following are equivalent:
$X$ satisfies the descending chain condition for closed subsets.
$X$ satisfies the ascending chain condition for open subsets.
Every subspace of $X$ is quasicompact.
Proof. Clearly, (1) and (2) are two different ways of saying the same thing. Assume (2). Let $Y$ be any subset of $X$, let $\mathfrak{U}$ be a family of open subsets of $X$, and suppose $Y \subseteq \bigcup_{U \in \mathfrak{U}} U$. Choose a well-ordering of $\mathfrak{U}$ and consider the sequence of partial unions: $U_0 \subseteq U_0 \cup U_1 \subseteq \cdots $ By the ascending chain condition, there is some $n$ such that $U_n = \bigcup_{U \in \mathfrak{U}} U$, thus we have the required finite subcover of $Y$.
Conversely, suppose every subset of $X$ is quasicompact. Consider an ascending chain of open subsets of $X$: $U_0 \subseteq U_1 \subseteq U_2 \subseteq \cdots$ By hypothesis, $U = \bigcup_n U_n$ is quasicompact, and $\{ U_0, U_1, U_2, \cdots \}$ is an open cover of $U$; but then there must be a finite subcover, so the ascending chain condition for open subsets of $X$ is satisfied. ◼
Condition (3) immediately implies:
Corollary. Any subspace of a noetherian topological space is again a noetherian topological space.
Proposition. Any noetherian topological space admits a decomposition into finitely many irreducible components.
Proof. Let $X$ be a noetherian topological space. Since $X$ satisfies the descending chain condition for closed subsets, the inclusion relation on the set of closed subsets of $X$ is well-founded, so we may apply well-founded induction. Clearly, the empty set is a union of finitely many irreducible closed subsets – after all, zero is a finite number. Suppose $Y$ is a closed subset such that every proper closed subset is a union of finitely many irreducible subsets. Then, either $Y$ is irreducible, or $Y = Z \cup W$ for two proper closed subsets $Z$ and $W$. By the induction hypothesis, $Z$ and $W$ is a union of finitely many irreducible subsets, so $Y$ is as well. By induction, every closed subset of $X$ is as a union of finitely many irreducible subsets – and $X$ is closed in $X$, so it too is a union of finitely many irreducible subsets.
It remains to be shown that $X$ can be written as a union of finitely many irreducible components. Consider the set of all irreducible subsets of $X$, partially ordered by inclusion. This is a chain-complete poset: indeed, if $Y_0 \subseteq Y_1 \subseteq Y_2 \subseteq \cdots$ is a chain of irreducible subsets of $X$, and $Y = \bigcup_n Y_n \subseteq Z \cup W$ for some closed subsets $Z$ and $W$ but $Y \nsubseteq W$, then for some $n$ we have $Y_n \nsubseteq W$; but by irreducibility $Y_m \subseteq Z$ for all $m \ge n$, so $Y \subseteq Z$, and therefore $Y$ is irreducible.
Moreover, any maximal irreducible subset is closed. Indeed, let $Y$ be any irreducible subset of $X$, and consider $\overline{Y}$. Let $U$ be an open set such that $U \cap \overline{Y}$ is non-empty; then $U \cap Y$ is also non-empty, since $Y$ is dense in $\overline{Y}$. Since $Y$ is irreducible, any closed set $Z$ such that $U \cap Y \subseteq Z \cap Y$ must satisfy $Z \cap Y = Y$. In particular, any closed set $Z$ such that $U \cap \overline{Y} \subseteq Z \cap \overline{Y}$ must have $Y \subseteq Z$, and so $\overline{Y} \subseteq Z$ too, as required. Thus, every irreducible subset of $X$ is contained in a maximal irreducible closed subset, a.k.a. an irreducible component. It follows that $X$ is a union of finitely many irreducible components. ◼