Does perfectly normal $\implies$ normal?

A question here on perfectly normal spaces got me into investigating the definition of such a space. The definition on wikipedia says

A perfectly normal space is a topological space X in which every two disjoint non-empty closed sets $E$ and $F$ can be precisely separated by a continuous function $f$ from $X$ to the real line $\mathbb{R}$: the preimages of $\{0\}$ and $\{1\}$ under $f$ are, respectively, $E$ and $F$. (In this definition, the real line can be replaced with the unit interval $[0,1]$.) It turns out that $X$ is perfectly normal if and only if $X$ is normal and every closed set is a $G_\delta$ set.

I took it for granted that a perfectly normal space was also assumed to be a normal space, which seems to be the case if you take the definition to be a perfect space where every closed set is $G_\delta$. If you take the other definition, I don't think you can assume automatically that the space is perfect. So this is kind of a follow up, from the first definition above, how can you conclude that a perfectly normal space is also normal?

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Related question: math.stackexchange.com/questions/72138/… –  Martin Sleziak Oct 13 '11 at 8:08
Let $F$ and $G$ be disjoint closed sets in a perfectly normal space $X$ as defined in Wikipedia. There is a continuous function $f:X\to [0,1]$ such that $F = f^{-1}[\{0\}]$ and $G = f^{-1}[\{1\}]$. Let $$U = f^{-1}\left[\left[0,\frac13\right)\right]$$ and $$V = f^{-1}\left[\left(\frac23,1\right]\right].$$ Since $f$ is continuous, $U$ and $V$ are open in $X$. Clearly $U \cap V = \varnothing$, $F\subseteq U$, and $G\subseteq V$, so $U$ and $V$ separate the closed sets $F$ and $G$. It follows that $X$ is normal.
One last thing to ask, do you know why $\mathbb{R}$ can be replaced by $[0,1]$ in this definition? –  Danielle Intal Oct 15 '11 at 4:51
@Danielle: The existence of separating functions from $X$ to $\mathbb{R}$ implies that $X$ is normal and that every closed set in $X$ is $f^{-1}[\{0\}]$ for some continuous $f:X\to\mathbb{R}$. If $F=f^{-1}[\{0\}]$, let $g(x)=\min\{|f(x)|,1\}$; $F=g^{-1}[\{0\}]$, and $g:X\to[0,1]$. Now suppose that $F,G$ are disjoint closed sets, $f,g:X\to[0,1]$ are continuous, $F=f^{-1}[\{0\}]$, and $G=g^{-1}[\{0\}]$. Let $h(x)=\dfrac{f(x)}{f(x)+g(x)}$; then $h:X\to[0,1]$ is continuous, $F=h^{-1}[\{0\}]$, and $G=h^{-1}[\{1\}]$. –  Brian M. Scott Oct 15 '11 at 9:44
I had that same $h$ in mind, but was unsure how to know $f(x)+g(x)\neq 0$ for some $x$ unless the codomain was $[0,1]$ instead of $\mathbb{R}$. Thanks for your comment! –  Danielle Intal Oct 15 '11 at 17:20