First some definitions:
A space $X$ is called completely regular if points and closed sets can be separated by functions (if $x\notin F$ with $F$ closed, there is a continuous $f:X\to[0,1]$ such that $f[F]=\{0\}$ and $f(x)=1$). (Hausdorff not assumed here.)
A space $X$ is called perfectly normal if disjoint closed sets can be precisely separated by functions (if $E$ and $F$ are disjoint closed sets, there is a continuous $f:X\to[0,1]$ such that $f^{-1}(0)=E$ and $f^{-1}(1)=F$). Equivalently (Vedenissoff's theorem), every closed set is a zero-set (= the set of zeros of some real valued continuous function). (Again, Hausdorff not assumed here.)
I can't find a reference for the following result anywhere:
Proposition: Every perfectly normal space is completely regular.
Note that the assumption cannot be weakened to normal spaces, or even to completely normal spaces. For example, the Sierpinski space is completely normal, but is not completely regular. It's not even regular.
I have provided a proof below; let me know if there is a problem or if you can find a reference.
Proof: Suppose $X$ is perfectly normal. Let $F$ be a closed set and $x$ a point not in $F$. By perfect normality, $F$ is the zero set of some continous $f:X\to[0,1]$. As $x\notin F$, we have $f(x)\ne 0$. It is then a simple matter to tweak $f$ to get $f(x)=1$.
For reference, this also follows from a more general result:
See this question for a proof.
Here, an $R_0$ space is a space whose Kolmogorov quotient is $T_1$. There are many equivalent characterizations (see wikipedia for details). One of them is that for a closed set $F$ and a point $x\notin F$, we have $F\cap\operatorname{cl}\{x\}=\emptyset$. A perfectly normal space has this property by an argument similar to the first proof above.