# Incorrect proof that normality is hereditary

I have just learned about normal spaces and that they are not hereditary, but I am having trouble seeing why. I understand why closed subspaces clearly inherit normality but I'm not sure why non-closed ones don't. What is wrong with the following proof?

Let $(X,\mathcal{T})$ be normal and let $A \subseteq X$ be a subspace. Let $C_1, C_2 \subseteq A$ be disjoint, nonempty closed subsets in the subspace topology. Then there exist two sets $D_1$ and $D_2$ which are closed in X such that $C_i = A\cap D_i$ for $i = 1,2$.

Since $X$ is normal, we have that there exists disjoint open sets, $U_1, U_2 \in X$ such that $D_1 \subset U_1$ and $D_2 \subset U_2$. Now we have that $C_i = (D_i \cap A) \subset (U_i \cap A)$. $U_i \cap A$ is open in the subspace topology. Since $U_1$ and $U_2$ are disjoint, so are $U_1 \cap A$ and $U_2 \cap A$.

Therefore, $A$ is normal.

What am I missing here? What's the flaw?

For disjoint open sets $U_1$ and $U_2$ to exist using the the condition of normality in $X$, the two closed sets $D_1$ and $D_2$ in $X$ must be disjoint to start with which is not guaranteed to be unless $A$ itself is a closed subspace.
It could happen that despite $\operatorname{cl}_A(C_1) \cap \operatorname{cl}_A(C_2) = C_1 \cap C_2 = \emptyset$, we have that $\operatorname{cl}_X(C_1) \cap \operatorname{cl}_X(C_2) \neq \emptyset$. If $D_i$ is such that it is closed and $D_i \cap A = C_i$, $i=1,2$, then $\operatorname{cl}_X(C_i) \subseteq D_i, i=1,2$ (as $D_i$ is closed in $X$ and contains $C_i$) and the $D_i$ would never be disjoint, and we cannot apply the normality in $X$.
To see this concretely look at the modification of the Tychonoff plank (a non-normal subspace of a normal space) I gave here. The $A, B \subseteq Z$ are disjoint and closed in $Z$ but their closures intersect in $(\infty_1, \infty_2)$ in the surrounding space $X \times Y$ (which is compact Hausdorff, so normal). I show that $A$ and $B$ cannot be separated by disjoint open sets in $Z$.