Is the lexicographic ordering on the unit square perfectly normal? It's known that order topology is completely normal, so the lexicographic ordering on the unit square is also completely normal. It's also known that the lexicographic ordering on the unit square is not metrizable. I am interested in whether it is perfectly normal. (A space is perfectly normal if for any two disjoint nonempty closed subsets, there is a continuous function $f$ to $[0,1]$ that "precisely separates" the two sets, meaning that the two closed sets are $f^{-1}(0)$ and $f^{-1}(1)$.) And how do we prove it?
 A: As Nate Eldredge points out in the comments, the book Counterexamples in Topology states that this space is not perfectly normal, but does not provide a proof. Here is the idea for a proof.
A perfectly normal space is a normal space in which every closed set is a $G_\delta$. So to prove this space is not perfectly normal, we'd like to find a closed set that is not a $G_\delta$. I claim that the subspace $C = [0,1] \times \{0,1\}$ works. ($C$ is the two horizontal lines on the top and bottom.) The reason is that if $U \subseteq [0,1] \times [0,1]$ is an open set containing $C$, then 
$$A_U = \{x \in [0,1] \ : \ \{x\} \times [0,1] \subseteq U\}$$
(i.e., the set of all vertical lines contained in $U$) is a co-countable subset of $[0,1]$. (This is because for every $p \in (0,1]$, $U$ must contain a neighborhood of $(p,1)$, and therefore $A_U$ must contain an interval of the form $(p-\varepsilon,p)$. Likewise, $U$ must contain a neighborhood of $(p,0)$, and therefore $A_U$ must contain an interval of the form $(p,p+\varepsilon)$. So every point of $[0,1] \setminus A_U$ is isolated.) Therefore any countable intersection of open neighborhoods of $C$ contains a horizontal line.
EDIT: Thanks to Nate Eldredge for helping me to simplify my original argument.
A: It is a known fact that any perfectly normal (countably) compact space $X$ is ccc. The proof is what you would try: start with an uncountable cellular family $\mathcal{U}$, choose a point $p_U \in U$ for each $U \in \mathcal{U}$ and consider the closed set $C=X \setminus \bigcup \mathcal{U}$. Since the space is perfectly normal, $C$ is a $G_\delta$ (say $C= \bigcap_nV_n$) and since $\mathcal{U}$ is uncountable you can find $n$ such that infinitely many of the $p_U$'s are outside of $V_n$. Now this infinite set of $p_U$´s has no limit points, contradicting the compactness of $X$.
Since $I^2_{lex}$ is compact and not ccc (both standard facts), it cannot be perfectly normal.
A: For a compact Hausdorff $X$ it is equivalent that $X$ is hereditarily Lindelöf or that $X$ is perfectly normal. Sketch: $X$ is hereditarily Lindelöf implies that every open set is an $F_\sigma$ (as $X$ is regular) so dually every closed set is a $G_\delta$. OTOH if $X$ is perfectly normal, every open set is an $F_\sigma$ so $\sigma$-compact and Lindelöf, making all open sets Lindelöf and $X$ hereditarily Lindelöf again. 
$[0,1]^2$ in the lexicographic order topology has a discrete subset $[0,1]\times \{\frac12\}$ so is definitely not hereditarily Lindelöf, so not perfectly normal either.
