[proofplan] We prove normality of the subspace directly from the closed-set separation definition. Given two disjoint closed subsets of $Y$, the [subspace topology](/page/Subspace%20Topology) realizes them as intersections of $Y$ with closed subsets of $X$; because $Y$ itself is closed in $X$, these subsets are closed in $X$. Normality of $X$ separates them by disjoint open subsets of $X$, and intersecting those open sets with $Y$ gives the required disjoint open neighborhoods in the subspace topology. If the adopted convention includes the $T_1$ axiom in the definition of normality, the $T_1$ property is inherited by subspaces, so the conclusion remains valid under that convention. [/proofplan] [step:Promote closed subsets of the subspace to closed subsets of the ambient space] Let $A \subset Y$ and $B \subset Y$ be closed subsets of $(Y,\tau_Y)$ such that $A \cap B = \varnothing$. Since $A$ is closed in the subspace topology, there exists a [closed set](/page/Closed%20Set) $F_A \subset X$ in $(X,\tau)$ such that \begin{align*} A = Y \cap F_A. \end{align*} Since $B$ is closed in the subspace topology, there exists a closed set $F_B \subset X$ in $(X,\tau)$ such that \begin{align*} B = Y \cap F_B. \end{align*} Because $Y$ is closed in $(X,\tau)$, both $Y \cap F_A$ and $Y \cap F_B$ are closed in $(X,\tau)$. Hence $A$ and $B$ are closed subsets of $X$. [guided] We start with the objects that must be separated in order to prove normality of $(Y,\tau_Y)$. Let $A \subset Y$ and $B \subset Y$ be closed in the subspace topology $\tau_Y$, and assume \begin{align*} A \cap B = \varnothing. \end{align*} The key point is that closedness in a subspace is not automatically the same as closedness in the ambient space. By the definition of the subspace topology, a subset of $Y$ is closed in $Y$ precisely when it is the intersection of $Y$ with a closed subset of $X$. Therefore there are closed subsets $F_A \subset X$ and $F_B \subset X$ such that \begin{align*} A = Y \cap F_A \end{align*} and \begin{align*} B = Y \cap F_B. \end{align*} Now we use the hypothesis that $Y$ is closed in $X$. Since $F_A$ is closed in $X$ and $Y$ is closed in $X$, their intersection $Y \cap F_A$ is closed in $X$. Thus $A$ is closed in $X$. The same argument applied to $F_B$ gives that $B = Y \cap F_B$ is closed in $X$. This is exactly where the closedness of $Y$ is needed: it allows closed subsets of the subspace $Y$ to be treated as closed subsets of the ambient normal space $X$. [/guided] [/step] [step:Separate the promoted closed sets in the normal ambient space] The sets $A$ and $B$ are disjoint closed subsets of the normal space $(X,\tau)$. By normality of $(X,\tau)$, there exist open sets $U,V \in \tau$ such that \begin{align*} A \subset U \end{align*} \begin{align*} B \subset V \end{align*} \begin{align*} U \cap V = \varnothing. \end{align*} [/step] [step:Restrict the separating neighborhoods to the subspace] Define subsets $U_Y \subset Y$ and $V_Y \subset Y$ by \begin{align*} U_Y := U \cap Y \end{align*} and \begin{align*} V_Y := V \cap Y. \end{align*} Because $U,V \in \tau$, the definition of $\tau_Y$ gives $U_Y,V_Y \in \tau_Y$. Since $A \subset Y$ and $A \subset U$, we have $A \subset U_Y$. Since $B \subset Y$ and $B \subset V$, we have $B \subset V_Y$. Finally, \begin{align*} U_Y \cap V_Y = (U \cap Y) \cap (V \cap Y) = (U \cap V) \cap Y = \varnothing. \end{align*} Thus every pair of disjoint closed subsets of $(Y,\tau_Y)$ can be separated by disjoint open subsets of $(Y,\tau_Y)$. [/step] [step:Verify the possible inherited $T_1$ requirement and conclude normality] If normality is defined only as separation of disjoint closed sets by disjoint open neighborhoods, the preceding step proves that $(Y,\tau_Y)$ is normal. If the convention also requires the $T_1$ axiom, then $(X,\tau)$ is $T_1$ because it is normal under that convention. Let $y \in Y$. Since $(X,\tau)$ is $T_1$, the singleton $\{y\}$ is closed in $X$. Therefore \begin{align*} \{y\} = Y \cap \{y\} \end{align*} is closed in $(Y,\tau_Y)$. Hence every singleton in $Y$ is closed, so $(Y,\tau_Y)$ is $T_1$. Combining this inherited $T_1$ property with the closed-set separation proved above, $(Y,\tau_Y)$ is normal under the convention that normal spaces are required to be $T_1$. [/step]