[proofplan] We prove the descending chain condition for closed subsets of $Y$. Given a descending chain of closed subsets in the subspace $Y$, each member is closed in $X$ because it is the intersection of $Y$ with a closed subset of $X$ and $Y$ itself is closed in $X$. The Noetherianity of $X$ then forces the same chain to stabilize, which is exactly the Noetherian condition for $Y$. [/proofplan] [step:Convert closed subsets of the subspace into closed subsets of the ambient space] Let $\mathbb{N}=\{1,2,3,\dots\}$. Let $(C_m)_{m\in\mathbb{N}}$ be a descending chain of closed subsets of $Y$ in the [subspace topology](/page/Subspace%20Topology), so \begin{align*} C_1 &\supset C_2 \supset C_3 \supset \cdots \end{align*} For each $m\in\mathbb{N}$, since $C_m$ is closed in the subspace $Y$, there exists a closed subset $F_m\subset X$ such that $C_m = Y\cap F_m$. Because $Y$ is closed in $X$ by hypothesis and $F_m$ is closed in $X$, their intersection $Y\cap F_m$ is closed in $X$. Hence each $C_m$ is closed in $X$. [guided] We start with the condition we must prove for $Y$: every descending chain of closed subsets of $Y$ stabilizes. Let $(C_m)_{m\in\mathbb N}$ be an arbitrary descending chain of closed subsets of $Y$, meaning $C_1 \supset C_2 \supset C_3 \supset \cdots$. The point is to compare this chain with closed subsets of $X$, because the Noetherian hypothesis is available only for $X$. The definition of the subspace topology says precisely that a subset is closed in $Y$ if and only if it is the intersection of $Y$ with a closed subset of $X$. Therefore, for each $m\in\mathbb N$, there exists a closed subset $F_m\subset X$ such that $C_m = Y\cap F_m$. Now we use the extra hypothesis that $Y$ is closed in $X$. Since both $Y$ and $F_m$ are closed subsets of $X$, their intersection is closed in $X$. Thus $C_m = Y\cap F_m$ is closed in $X$ for every $m\in\mathbb N$. This is the only place where the assumption that $Y$ is closed in $X$ is needed. [/guided] [/step] [step:Apply the Noetherian condition in the ambient space] The same sequence $(C_m)_{m\in\mathbb N}$ is now a descending chain of closed subsets of $X$. Since $X$ satisfies the descending chain condition for closed subsets, there exists $N\in\mathbb N$ such that $C_m = C_N$ for every $m\ge N$. Therefore every descending chain of closed subsets of $Y$ stabilizes. Hence $Y$, with the subspace topology, is Noetherian. [guided] From the previous step, each $C_m$ is closed not only in the subspace $Y$ but also in the ambient space $X$. The inclusions $C_1 \supset C_2 \supset C_3 \supset \cdots$ are therefore a descending chain of closed subsets of $X$. By the hypothesis that $X$ satisfies the descending chain condition for closed subsets, this chain stabilizes. Thus there exists an index $N\in\mathbb N$ such that $C_m = C_N$ for every $m\ge N$. This is exactly the stabilization condition required for the original arbitrary descending chain of closed subsets of $Y$. Since the chain in $Y$ was arbitrary, every descending chain of closed subsets of $Y$ stabilizes, so $Y$ with the subspace topology is Noetherian. [/guided] [/step]