for perfect normality it is not sufficient that the topological space be $T_4$ and every point in it be a $G_{\delta}$-set So there is a part of a problem on the book General Topology by Willard (15C-part 3) that says

it is not sufficient for perfect normality that $X$ be $T_4$ and every point in $X$ be a $G_{\delta}$-set (countable intersections of open sets).

I've been trying to find an example for this but I didn't have much luck. Any hint or answers are appreciated.
 A: Let $X=[0,1] \times [0,1]$ in the order topology induced from the lexicographic order $(x,y) \le (x',y')$ iff ($x < x'$) or ($x=x'$ and $(y \le y')$). As all order topologies are $T_5$, so is this one. It's also first countable, so singletons are $G_\delta$ too. As it's compact too, it cannot be perfectly normal as that would imply $X$ is hereditarily Lindelöf, which it is not, $[0,1] \times \{\frac12\}$ being closed, discrete and uncountable.
A: Let $X$ be the space of countable ordinals, Willard’s $\Omega$, with the usual order topology. All linearly ordered spaces with the order topology are $T_5$ (completely/hereditarily normal and $T_1$), and $X$ is easily seen to be first countable, so singletons are $G_\delta$ sets. Let $C$ be the set of limit ordinals in $X$; clearly $C$ is closed.
Let $U$ be an open nbhd of $C$. Then there is a function $\varphi_U:C\to X$ such that $(\varphi_U(\alpha),\alpha]\subseteq U$ for each $\alpha\in C$. The pressing-down lemma implies that there are an $\eta_U\in X$ and a cofinal $C_U\subseteq C$ such that $\varphi_U(\alpha)=\eta_U$ for each $\alpha\in C_U$.
Now suppose that $\mathscr{U}=\{U_n:n\in\Bbb N\}$ is a family of open nbhds of $C$. Let $\eta=\sup_{n\in\Bbb N}\eta_{U_n}$, and show that
$$\bigcap\mathscr{U}\supseteq(\eta,\omega_1)\supsetneqq C\,.$$
