Sorgenfrey plane is not normal, questions about the proof I have some questions about the proof that I saw on that page: 
https://dantopology.wordpress.com/2009/10/01/a-short-note-about-the-sorgenfrey-line/.


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*Why $H_0$ and $H_1$ are closed?

*"Since $\mathbb{P}$ is not an $F_\sigma$ subset of $\mathbb{R}$, there exists $z \in \mathbb{Q}$ and there exists an $n$ such that $z$ is in the closure of $P_n$ in the usual topology of $\mathbb{R}$." - Why we can say that?

 A: For your first question, recall that a set in closed if its complement is open. From the definition of $H_0$ and $H_1$, we can see that they are in the diagonal $y=-x$ of $\mathbb{R}^2$.
Choose a point $(x,y) \in S \times S - H_i$ ($i=0$ or $i=1$), take an $\epsilon > 0$ small enough so that $[x,x+\epsilon) \times [y,y+\epsilon)$ does not intersect the diagonal, and hence does not intersect $H_0$ or $H_1$. In fact, if $(x,y)$ is "above" the diagonal, you don't really care about the $\epsilon$ and if $(x,y)$ is on the diagonal, the fact that you can take the intervals to be closed on the right side is what makes it work.
Therefore, $S \times S - H_i$ is a neighbourhood of all its points, so is open. Hence, $H_i$ is closed.
Regarding you second question, let's consider what happens when we suppose the contrary (I personally like this way when I don't understand such a statement). Suppose that for every $z\in \mathbb{Q}$ and every $n\in\mathbb{N}$, $z$ is not in the closure of $P_n$. I then claim that every $P_n$ is closed: the closure of $P_n$ is included in $\mathbb{P}$ by assumption, if there were a $p\in\mathbb{P}$ in the closure of $P_n$ but not in $P_n$, then, by density, we could find a $q\in\mathbb{Q}$ also in the closure.
But, as stated before in the article, $\mathbb{P} = \cup P_n$ would be a countable union of closed subsets, ie a $F_{\sigma}$ subset.
