What's wrong with the following false-proof that a subspace of a normal space is normal?

Question

The fact that a subspace of a normal topological space is normal is known to be wrong.
I'm trying to find the wrong step in the following proof:

Let $X$ be a normal topological space and $Y$ be a subspace. Let $C_1, C_2$ be closed sets in $Y$.
So there exist $C_1', C_2'$ closed in $X$ such that $C_i = Y \cap C_i' $ for $i=1,2$.
Since $X$ is normal, there are open disjoint sets $U_1, U_2$ such that $C_i' \subset U_i$.
Thus $C_i=C_i' \cap Y \subset U_i \cap Y$.
So if we let $V_i = U_i \cap Y$, we get that $V_1, V_2$ are open in $Y$, disjoint, and satisfy $C_i \cap V_i$.
Therefore $Y$ is a normal topological space.
Where is the mistake?

Answer 1

Since $X$ is normal, there are open disjoint subsets $U_1, U_2$ such that $C_i' \subset U_i$

No, this only follows if you also know that $C_1' \cap C_2' = \varnothing$!

This is actually the source of two separate problems in your proof. Can you identify them?