A quotient topology of Niemytzki Plane

Let $X$ is the classical Niemytzki plane. We consider the points of the line $y=0$. We paste all the points of $Q$ to one point, and paste all the other points of the line $y=0$ to the other point. Thus we generates a new topological space. It's the quotient topology of Niemytzki Plane. Now my question is this:

Does this new space is still regular? (We know the Niemytzki Plane is a Tychonoff space.)

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This seems to be Example 14.5 in Willard's General Topology, p.93. – Martin Sleziak Jul 21 '12 at 12:08
What is $Q{}{}$? – Chris Eagle Jul 21 '12 at 12:09
@Chris Eagle $Q$ denotes all the rational numbers. – Paul Jul 21 '12 at 12:21
@Martin Sleziak A little pity.The page 93 of the book cann't be seen on google:) – Paul Jul 21 '12 at 12:25
93, 94 – Martin Sleziak Jul 21 '12 at 12:32

It is not even Hausdorff. If $p$ is the point resulting from the identification of$P=\{\langle x,0\rangle:x\in\Bbb R\setminus\Bbb Q\}$, and $q$ is the point resulting from the identification of $Q=\{\langle x,0\rangle:x\in\Bbb Q\}$, then $p$ and $q$ cannot be separated by disjoint open sets in the new space. This follows from the fact that $P$ and $Q$ cannot be separated by disjoint open sets in the Niemytzki plane, as is shown here. (It can also be shown using the Baire category theorem.)
@Martin: Yes, that’s essentially the argument that I had in mind. It doesn’t apply the BCT to the Niemytzki plane, though: it uses the fact that $\Bbb R\setminus\Bbb Q$ is a Baire space in the usual topology. – Brian M. Scott Jul 22 '12 at 8:05
Thanks Brain for your answer. Do you mean in your comment that we apply the BCT not to the whole spae, but only to the line $y=0$ with usual topology? – Paul Jul 22 '12 at 11:52