Show that the Scattered Line (also called the Michael Line) is $T_4$ I just finished reading through the section about normal spaces (resp. $T_4$-spaces) from Willard General Topology text, and came across the following problem:

Show that the Scattered Line (also called the Michael Line) is $T_4$.

I'm not sure how to prove this, let alone get started on the problem. Any help would be much appreciated.
Note: The Scattered Line (also called the Michael Line) is given as follows: Let $\mathbb{R}$ be the real number line. Let $\mathbb{P}$ be the set of all irrational numbers. Let $\mathbb{Q}=\mathbb{R}-\mathbb{P}$, the set of all rational numbers. Let $\tau$ be the usual topology of the real line $\mathbb{R}$. The following is a base that defines a topology on $\mathbb{R}$.
$$\mathcal{B}=\tau \cup \left\{\left\{ x \right\}: x \in \mathbb{P}\right\}$$
The real line with the topology generated by $\mathcal{B}$ is called the Scattered Line (also called the Michael Line).
 A: A LOTS (linearly ordered topological space) is a triple $\langle X,\tau,\le\rangle$, where $\le$ is a linear order on $X$, and $\tau$ is the associated order topology. The desired result follows quite easily from an important result about thise spaces: if $\langle X,\tau,\le\rangle$ is a LOTS, then $X$ is $T_5$ (i.e., hereditarily normal and $T_1$). I posted a proof of this result here.
Let $X=(\Bbb P\times\Bbb Z)\cup(\Bbb Q\times\{0\})$, and give $X$ the lexicographic order topology. Let $M$ be the subspace $\Bbb R\times\{0\}$ of $X$; then $M$ is homeomorphic to the Michael line by the map $\langle x,0\rangle\mapsto x$. Thus, the Michael line is a subspace of a LOTS and as such is normal.
A: This can be viewed (as Brian M. Scott also does) from a more general point of view:
A $T_1$ space is called monotonically normal iff there is a so-called $\mu$-operator that assigns an open subset $\mu(x,O)$ for every pair $x,O$ where $x \in O$ and $O$ is open in $X$ and so that

*

*$x \in \mu(x,O) \subseteq O$ and

*Whenever we have $\mu(x,U) \cap \mu(y,V) \neq \emptyset$ we can conclude that $x \in V$ or $y \in U$.

This looks rather innocent but is in fact quite powerful: it's a hereditary property: if $X$ is monotonically normal and $A \subseteq X$ then so is $A$ in its subspace topology: if $O \cap A$ is an open subset of $A$ (so $O$ is open in $X$) we define $\mu_A(x, O \cap A) = \mu(x,O) \cap A$, when $\mu$ is the operator on $X$.
Fact

If $X$ is monotonically normal, then $X$ is $T_5$ (aka hereditarily normal /completely normal).

Proof: as being monotonically normal is hereditary, it suffices to show $X$ is normal (the $T_1$ is part of the definition). So let $A$ and $B$ be disjoint closed sets of $X$.
Define $$U = \bigcup \{\mu(a, X\setminus B)\mid a \in A\}, V = \bigcup \{\mu(b, X\setminus A)\mid b \in B\}$$
and note that $A \subseteq U$, $B \subseteq V$ and these sets are open. They are also disjoint, for if there not, there must be some $a_0 \in A$ and some $b_0 \in B$ such that $\mu(a_0, X\setminus B) \cap \mu(b_0, X\setminus A) \neq \emptyset$ and by property 2 we'd then have $a_0 \in x \setminus A$ or $b_0 \in X\setminus B$, both of which are absurd. So that completes the proof.
Useful fact:

It suffices to define $\mu(x,O)$ only for $O$ from some given fixed base $\mathcal{B}$ for $X$ and all $x \in O$, having the right properties. Then defining $\mu(x, \bigcup_{i \in I} O_i) = \bigcup \{\mu(x,O_i)\mid x \in O_i, i \in I\}$, where all $O_i \in \mathcal{B}$, defines a $\mu$-operator for all open subsets.

The proof is an easy exercise in the definitions.
Some examples: Any metric space is monotonically normal, we can define $\mu$ on a base by $\mu(x, B(x,r)) = B(x, \frac{r}{2})$, and the triangle inequality takes care of defining property 2.
The Sorgenfrey line is monotonically normal: define $\mu(x, [a,b)) = [x,b)$ on base elements, and checking property 2 is a simple matter of case checking.
So we then have "for free" that metric spaces are $T_5$ and so is the Sorgenfrey line. In fact all ordered spaces are monotonically normal too, this has a longer proof that I won't give here now.
But back to the Michael Line, or Scattered line: this is a special case of a more general construction: if $(X, \mathcal{T}_X)$ is a space and $A \subseteq X$ a subspace of $X$, then $X_A$ is the space with point set $X$ and topology generated by $\mathcal{T}_X \cup \mathscr{P}(X\setminus A)$; in effect: $A$ keeps its original subspace topology and all points outside of $A$ become isolated points.
The Michael line is just $\Bbb R_{\Bbb Q}$, where $\Bbb R$ has its usual topology.
Proposition

If $X$ is monotonically normal and $A \subseteq X$ then $X_A$ is also monotonically normal

Proof: $T_1$ is trivial, as we have a finer topology.
Let $\mu$ be the operator on $X$. Then a $\mu$ operator on the generating set of $X_A$ is defined by $\mu'(x,O)=\mu(x,O)$ if $O \in \mathcal{T}_X$ and $\mu'(x,B)=\{x\}$ if $B \subseteq X\setminus A$. This works, as is quite easy to check.
Corollary

The Michael line is monotonically normal and hence hereditarily normal.

Many spaces with good high separation properties have it for the reason that they're monotonically normal. No longer a very fashionable notion/property, but I quite like it.
Hope this helps someone.
