I want to first point out that this question is not intended to be marked as a duplicate of Every open subspace of $X$ is normal implies that $X$ is completely normal. Instead it is an extension of that question, since there is something about the post that I don't quite understand.
The answer given in the question link provided goes as follows:
Suppose $A$ and $B$ are completely separated in $X$, i.e. $\operatorname{cl}_X(A) \cap B = A \cap \operatorname{cl}_X(B) = \emptyset$. If we'd know their closures were disjoint (which we don't!) we could use normality even in $X$ (separating the closures), but we cannot.
So define $Y = X \setminus (\operatorname{cl}_X(A) \cap \operatorname{cl}_X(B))$, cutting out the potential problem set. Then $Y$ is open in $X$, and so $Y$ is normal by assumption. Also $\operatorname{cl}_Y(A)$ and $\operatorname{cl}_Y(B)$ are disjoint in $Y$ (check this !). This can be shown using the general fact that $\operatorname{cl}_Y(Z) = \operatorname{cl}_X(Z) \cap Y$ for all subsets $Z$ of $X$, and $A$ and $B$ being completely separated in $X$. Now apply normality of $Y$ to these closures in $Y$ to finish the proof.
The issue I have with this proof is that I would think we would need to start with any open subspace $Y$ of $X$, suppose it's normal, and then arrive at the fact that $X$ is completely normal. But to say that $Y= X \setminus (\operatorname{cl}_X(A) \cap \operatorname{cl}_X(B))$ would appear to take away the arbitrariness of $Y$. In other words, it's not obvious to me that any open subspace $Y$ of $X$ can be written as $X \setminus (\operatorname{cl}_X(A) \cap \operatorname{cl}_X(B))$. Perhaps someone can elucidate this to me. Thanks in advance.
You have to prove that any two completely separated subsets $A, B$ of $X$ have disjoint open neighborhoods. To this end one considers the open subspace $Y$ of $X$. By assumption it is normal, and this allows to find disjoint open neighborhoods of $A, B$ in $Y$. These are also open in $X$ which finishes the proof. In this proof it is irrelevant what the origin of $Y$ is, we only need the fact that it is open.
In fact, it would be sufficient to require that all open subspaces $Y$ of the above special form are normal.