Subspaces of normal spaces aren't always normal An exercise I'm working on asks to identify the error in the below proof.
(False) Statement: Prove that normality is hereditary.
Proof.
Suppose $X$ is normal with $Y \subset X$. Let $E$, $F$ be closed in $X$ with $E \cap F = \varnothing$. Then
$$G = E \cap Y \hspace{1cm} H = F \cap Y$$
are closed in $Y$. Since $X$ is normal, there exist $X$-open sets, $U$ and $V$ so that
$$E \subset U \hspace{0.5cm} F \subset V \hspace{0.5cm} U \cap V = \varnothing$$
Then $U \cap Y$ and $V \cap Y$ are open in $X$. Moreover,
$$G \subset U \cap Y \hspace{1cm} H \subset V \cap Y$$
and so $Y$ is normal. Thus, normality is hereditary.

I cannot see the error with the proof given here. Any advice on what goes wrong?
 A: Normality is only preserved bij closed subspaces. To see this, let $(X,\mathcal{T})$ be a normal topological space and let $Y \subseteq X$ be closed, where $(Y,\mathcal{T}|_Y)$ is equipped with the subspace topology.
Let $A,B \subseteq Y$ be closed in $Y$, then they are also closed in $X$. Hence open sets $U,V \subseteq X$ exist such that $A \subseteq U$, $B \subseteq V$ and $U \cap V = \emptyset$. But then $U \cap Y$ and $V \cap Y$ are open sets in $Y$ still containing $A$ and $B$ respectively, and of course $(U \cap Y)\cap(V \cap Y) = \emptyset$. So $(Y,\mathcal{T}|_Y)$ is normal.
The fact that this does NOT carry over to open sets, can easily be disproved by a counterexample.
Let $X = \{a,b,c,d\}$ and $\mathcal{T} = \{ \emptyset,\{d\},\{b,d\},\{c,d\},\{b,c,d\},X\}$. Then $(X,\mathcal{T})$ is a topology, and because it has no pair of disjoint non-empty closed subsets, it is ny default normal. Consider $Y = \{b,c,d\}$ however (and mark that this is an open subset of $X$).Then $\mathcal{T}|_Y = \{\emptyset,\{d\},\{b,d\},\{c,d\},Y\}$ in which $\{b\}$ and $\{c\}$ are disjoint closed sets in $(Y,\mathcal{T}|_Y)$ that cannot be separated by disjoint open sets in $(Y,\mathcal{T}|_Y)$. Hence $(Y,\mathcal{T}|_Y)$ is not normal.
A: I think the error is in the beginning when you pick G and H. As I read this, you are considering some arbitrary closed subsets of $X$, say $E$ and $F$ where $E\cap F=\emptyset$ and then considering $G=E\cap Y$ and $H=F\cap Y$. Note, that in order to show that $Y$ is normal, you must consider arbitrary closed sets $G$ and $H$ in $Y$ such that $G\cap H=\emptyset$. Then, we can say that there are closed sets $E$ and $F$ in $X$ such that $G=E\cap Y$ and $H=F\cap Y$, but this does not imply that $E\cap F=\emptyset$. The fact that $E\cap F=\emptyset$ is key to your argument here, and I believe is the flaw.
