Engelking states a theorem(2.1.7) in the book General Topology as follows:
For every $T_1$ spaces the following are equivalent:
(1) Every subspace of $X$ is normal.
(2) Every open subspace of $X$ is normal.
(3) Two separated sets have disjoint open neighborhoods.
I was asked to show that (1) and (3) are equivalent, however, I don’t know where the $T_1$ condition is used in the proof. My proof goes as follows:
If $X$ satisfies (3), then let $Y$ be a subspace and $A,B\subset Y$ disjoint closed sets in the subspace $Y$. Denote the closure operator in $X$ by $Cl$ and in $Y$ by $Cl^*$. We claim that $A, B$ are separated in $X$: $$Cl(A)\cap B= Cl(A)\cap Cl^*(B) = Cl(A)\cap Cl(B)\cap Y = Cl^*(A)\cap Cl^*(B)$$ But this is empty by the fact that $A, B$ are closed. Then we can take disjoint open neighborhood of $A,B$ in $X$, and intersect both by $Y$.
If $X$ satisfies (1), then take two separated sets $A, B$ in $X$. The open subspace $Y=X-(Cl(A)\cap Cl(B))$ contains each of $A $ and $B$ by the fact that they are separated. The closure of them in $Y$ is empty: $Cl^*(A)\cap Cl^*(B)= Cl(A)\cap Cl(B)\cap Y$ is empty because $Y$ does not contain $Cl(A)\cap Cl(B)$. Take disjoint open neighborhood of $Cl^*(A)$ and $Cl^*(B)$ in $Y$, then by the fact that $Y$ is open, we are done with the proof.
So where exactly did I use $T_1$ condition in my proof?