Perfectly Normal is hereditary The definitions I'm working with:

$(X, T )$ is called perfectly normal if whenever $C$ and $D$ are disjoint, nonempty, closed subsets of $X$, there exists a continuous function $f : X \rightarrow [0, 1]$ s.t $C = f^{−1}(\{0\})$ and $D = f^{−1}(\{1\})$.
A topological space is normal if and only if every pair
of disjoint, nonempty closed sets can be separated by a continuous function

In particular the definition of normal doesn't assume $T_1$. I can prove that every perfectly normal space is normal, but since normal is not hereditary I don't see how that helps. My problem is that any two disjoint closed sets in a subspace are not necessarily disjoint in X.
I have also tried working with the alternative definition of perfectly normal, i.e a space is perfectly normal iff it is normal and every closed subset is $G_{\delta}$, but I run into the same problem namely that the closed disjoint sets of the subspace are not necessarily disjoint in X. Any help would be appreciated.
 A: HINT: Let $Y$ be a subspace of $X$. First show that if $F$ is a relatively closed subset of $Y$, there is a continuous $f:Y\to[0,1]$ such that $F=f^{-1}[\{0\}]$. Then let $F$ and $G$ be disjoint relatively closed subsets of $Y$. By the first result there are continuous $f,g:Y\to[0,1]$ such that $F=f^{-1}[\{0\}]$ and $G=g^{-1}[\{0\}]$. Consider the function
$$h:Y\to[0,1]:y\mapsto\frac{f(y)}{f(y)+g(y)}\;.$$
A: Note that your definition of perfectly normal

$(X, T )$ is called perfectly normal if whenever $C$ and $D$ are disjoint, nonempty, closed subsets of $X$, there exists a continuous function $f : X \rightarrow [0, 1]$ s.t $C = f^{−1}(\{0\})$ and $D = f^{−1}(\{1\})$.

is incorrect.  You need to remove the "nonempty" from there.
The condition must be satisfied for all disjoint closed sets $C$ and $D$, even if one of them is empty.  If one of them is empty, say $D$, the condition says that each closed set $C$ must be the zero set of some real valued function (which implies each closed set is a $G_\delta$).
If you keep the nonempty in the definition, that will not be the case: each ultraconnected space would trivially be perfectly normal, for the trivial reason that there are no disjoint nonempty closed subsets; but that is not true.  For example in the Sierpinski space $X=\{0,1\}$ with $1$ as the only open point, the singleton $\{0\}$ is closed, but is not a $G_\delta$.
