Proof for "every subset of a completely normal space is normal". Let $X^{*}$ be a subset of the completely normal space $X$. Let $A$ and $B$ be two disjoint closed subsets in $X^{*}$. $c$ is the closure operator in $X$, and $c^{*}$ is the closure operator in $X^*$
My book says "$A\cap c(B)=c^* (A)\cap c(B)=X^*\cap c(A)\cap c(B)=c^*(A)\cap c^*(B)=\emptyset$. Similarly, $c(A)\cap B=\emptyset$."


*

*How is $X^*\cap c(A)\cap c(B)=c^*(A)\cap c^*(B)$? Isn't this wrong?

*My proof is $A\cap c(B)=c^* (A)\cap c(B)=X^*\cap c(A)\cap c(B)=c(A)\cap c^*(B)=c(A)\cap B$. As $A$ and $B$ are disjoint sets in $X$, $c(A)\cap B=A\cap c(B)=\emptyset$. This proves that $A$ and $B$ are separable in $X$, and are hence contained in disjoint open sets as $X$ is completely normal). Let these open sets be $A'$ and $B'$. Then $X^*\cap A'$ and $X^*\cap B'$ will be disjoint open sets in $X^*$ containing the disjoint closed sets $A$ and $B$, making $X^*$ normal. 
Is this argument coherent?
EDIT- A completely normal space, as defined in my book, is one in which any two separated subsets are contained in disjoint open sets. 
 A: You always have $c^*(A)=X^*\cap c(A)$ if $A$ is a subset of $X^*$. Then $X^*\cap c(A)\cap c(B)=(X^*\cap c(A))\cap (X^*\cap c(B))=c^*(A)\cap c^*(B)$.
But this could even be improved:
Note that $A\cap c(B)=X^*\cap A\cap c(B)=A\cap c^*(B)$ which shows that a subspace of a completely normal space is completely normal.
The reverse is also true: If each subspace is normal, then $X$ is completely normal.
I've spoiler-protected the solution. Hover over the box to see it.

 Let $A$ and $B$ be two non-empty separated subsets of $X$. Use the normality of the subspace $X-(\text{cl}A\cap\text{cl}B)$.

A: As to 1: 
$A\cap c(B)=c^\ast (A)\cap c(B)$ ($A = c^\ast(A)$ as $A$ is closed in $X^\ast$) 
$ \ldots = (X^\ast \cap c(A)) \cap c(B)$ (by the general formula $c^\ast(S) = c(S) \cap X^\ast$ for all subsets $S$ of $X^\ast$)
$= (X^\ast \cap c(A)) \cap (X^\ast \cap c(B)) = c^\ast(A) \cap c^\ast(B)$ (same formula again) 
$= A \cap B =\emptyset$, as $A$ and $B$ are closed and disjoint (in $X^\ast$).  
Completely symmetrically we have that $c(A) \cap B = \emptyset$ as well, so $A$ and $B$ are separated sets in $X$. 
As to 2.: indeed taking the intersection with $X^\ast$ of the two disjoint open sets around $A$ and $B$, that must exist by being completely normal, are clearly disjoint open sets around $A$ and $B$ in $X^\ast$, basically by definition of the subspace topology (and disjoint sets stay disjoint when we make them smaller....). 
So the key observation is that disjoint closed sets in some subspace are always separated sets in the larger space.
For the reverse (every subspace normal implies completely normal) we take two separated sets $A$ and $B$, and take the subspace $X^\ast = X \setminus (c(A) \cap c(B))$ (an open subspace of $X$), and we show that $A \subset X^\ast, B \subset X^\ast$, and $A$ and $B$ are closed and disjoint in $X^\ast$. The separating open sets are still open in $X$ (as we work in an open subspace). Note that we "only" need that all open subspaces of $X$ are normal.
