Is $T_1$ condition necessary in the definition of completely normality?

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?

• Does author require normal spaces to be Hausdorff? – William Elliot Dec 14 '18 at 3:31
• @WilliamElliot The author requires normal spaces to be both $T_1$ and $T_4$. I understand it now-I use this book just for a reference so I was not aware it has an alternate definition for normal space. Thank you for the comment. – William Sun Dec 14 '18 at 3:35

Be aware that Engelking has the tendency to assume extra separation axioms in some definitions: compact/paracompact includes Hausdorff, normal and regular includes $$T_1$$, perfectly normal includes normal (which includes $$T_1$$) etc.
So in order to have a space all of whose subspaces are normal, we can only consider $$T_1$$ spaces to begin with. It's the price of admission, as it were.