A direct proof that this space is normal I recently came across a question asking to show that by adding a single point to the Moore plane one can obtain a normal space. Instead of just going for the one-point compactification, I constructed a different space, which I can show is normal by showing it is paracompact and Hausdorff. However I would like a more direct proof of its normality.
Let $\mathbb M = \{ (x,y) \in \mathbb R^2 : y \geq 0 \}$, and consider the subsets $\mathbb M_0 = \{ (x,y) \in \mathbb M : y = 0 \}$ and $\mathbb M_1 = \{ (x,y) \in \mathbb M : y > 0 \}$. For each $\mathbf x \in \mathbb R^2$ and $r > 0$ let $B_r ( \mathbf x )$ be the usual open disc of radius $r$ centered at $\mathbf x$. The Moore plane is the space obtained by defining the following sets to be the basic open neighborhoods of each $\mathbf x = (x,y) \in \mathbb M$:


*

*If $\mathbf x \in \mathbb M_1$, the basic open neighborhoods of $\mathbf x$ are the $B_r ( \mathbf x ) \cap \mathbb M$ for $r > 0$.

*If $\mathbf x \in \mathbb M_0$, the basic open neighborhoods of $\mathbf x$ are the $D_r ( \mathbf x ) := \{ \mathbf x \} \cup B_r ( (x,r) )$ for $r > 0$.


My space $X = \mathbf M \cup \{ \star \}$ is given the topology where


*

*Every open subset of $\mathbb M$ is open in $X$.

*The basic open neighborhoods of $\star$ are of the form $\{ \star \} \cup \bigcup \{ D_{r ( \mathbf x )} ( \mathbf x ) : \mathbf x \in Z \}$ where $Z$ is a cofinite subset of $\mathbb M_0$, and $r( \mathbf x ) > 0$ for each $\mathbf x \in Z$.


Can someone provide an outline of a direct proof (using either the "disjoint closed sets" or "open set containing a closed set" definition) that this space is normal?

I will add a fairly complete outline of one part of the proof of normality. (Thanks to Ire Shaw for helping me focus on this case at least.)
Suppose that $F,E$ are disjoints closed subsets of $X$, neither of which contains $\star$. It follows that both $F \cap \mathbb M_0$ and $E \cap \mathbb M_0$ are finite, so we may enumerate them, respectfully, as $$\{ \mathbf x_1 = (x_1,0) , \ldots , \mathbf x_n = (x_n,0) \}; \quad \{ \mathbf u_1 = (u_1,0) , \ldots , \mathbf u_m = (u_m,0) \}.$$
For $i \leq n$ there is an $r^\prime_i > 0$ such that $D_{r^\prime_i} ( \mathbf x_i ) \cap E = \emptyset$. Simiarly, for $j \leq m$ there is an $s^\prime_j > 0$ such that $D_{s^\prime_j} ( \mathbf u_j ) \cap F = \emptyset$.
For $i \leq n$ define $$r_i = \min \left( \left\{ \frac{r^\prime_i}2 \right\} \cup \left\{ \frac{(x_i-u_j)^2}{4s^\prime_j} : j \leq m \right\} \right)$$
For $j \leq m$ define $$s_j = \min \left( \left\{ \frac{s^\prime_j}2 \right\} \cup \left\{ \frac{(x_i-u_j)^2}{4r^\prime_i} : i \leq n \right\} \right)$$
One can show that for $i \leq n$ and $j \leq m$ we have $D_{r_i} ( \mathbf x_i ) \cap D_{s_j} ( \mathbf u_j ) = \emptyset$.
For $\mathbf{x} \in F \cap \mathbb M_1$ there is an $r^\prime(\mathbf x)$ such that $B_{r^\prime(\mathbf x)} ( \mathbf x ) \cap E = \emptyset$. It can be shown that for $j \leq m$ since $\mathbf x \notin D_{s^\prime_j} ( \mathbf u_j )$, then $\mathbf x \notin \overline{ D_{s_j} ( \mathbf u_j ) }$, and so there is an $r_j ( \mathbf x ) > 0$ such that $B_{r_j} ( \mathbf x ) \cap D_{s_j} ( \mathbf u_j ) = \emptyset$. Set $$r ( \mathbf x ) = \min \left( \left\{ \frac{r^\prime ( \mathbf x )}{2} \right\} \cup \left\{ r_i ( \mathbf x ) , \ldots , r_m ( \mathbf x ) \right\} \right).$$
Do the same for $\mathbf u \in E \cap \mathbb M_1$ to get an $s ( \mathbf u ) > 0$ such that


*

*$B_{2 \cdot s ( \mathbf u )} ( \mathbf u ) \cap F = \emptyset$;

*$B_{s ( \mathbf u ) } ( \mathbf u ) \cap D_{r_i} ( \mathbf x_i ) = \emptyset$ for $i \leq n$.


Set 


*

*$U = \bigcup_{i \leq n} D_{r_i} ( \mathbf x_i ) \cup \bigcup_{\mathbf x \in F \cap \mathbb M_1} B_{r ( \mathbf x )} ( \mathbf x )$

*$V = \bigcup_{j \leq m} D_{s_j} ( \mathbf u_j ) \cup \bigcup_{\mathbf u \in E \cap \mathbb M_1} B_{s ( \mathbf u )} ( \mathbf u )$


Then $F \subseteq U$, $E \subseteq V$, and $U \cap B = \emptyset$.

The real difficulty appears to be just showing regularity at $\star$. That is, if $F \subseteq X$ is closed and does not contain $\star$, then there are open sets $U,V$ such that $\star \in U$, $F \subseteq V$ and $U \cap V = \emptyset$.
I can achieve this if $F \subseteq \mathbb M_0$, however I cannot come to grips on how to handle points in $F \cap \mathbb M_1$.
 A: Choose any pair of close sets $A,B$.
If both $A$ and $B$ do not contain the point $⋆$, then $A^c$, $B^c$ are both open neighborhoods of $⋆$. Hence $A$, $B$ contains only finitely points of $\mathbb{M}_0$.
Note that the restriction of $X$ to the upper halfplane $\mathbb{M}_1$ is equiped with the normal topology on $\mathbb{R}^2$.Then if we write $A=A_0\cup A_1$ where $A_i=A\cap M_i$,$i=1,2$, and similarily $B=B_0\cup B_1$, easily $A_1,B_1$ can be seperated by two open sets $V_1,U_1$.
Since $A_0$, $B_0$ contains only finitely points, write $A_0= \{ a_1,\dots a_n\}$ and $B_0=\{ b_0 \dots b_m \}$. There exist a $R>0$ such that for any $r<R$ we have:
$$\cup D_r(a_i) \bigcap \cup D_r(b_i)=\emptyset$$
If for all $r$ the equality $(\cup D_r(a_i)) \cap U_1 \not = \emptyset$ holds, it will follows that there exist at least a $a_k$ belongs to $B$. Similarily the argument goes for $A$.
Then for a proper $r$ we find that the sets $V = V_1 \cup (\cup D_r(a_i))$ and $U= U_1 \cup (\cup D_r(b_i))$ are the desired ones.
If there exist a set in $A$ and $B$ which contains $*$, says $*\in A$. Since $A\cap B =\emptyset$, $* \not\in B$, that is $B^c$ is a open neighborhood of $*$, then $B$ contains only finitely points of $\mathbb{M}_0$. 
Write $B_0 = \{ b_1,\dots,b_k \}$. First we want to find open neighbourhoods of $A_0 =  A \cap \mathbb{M}_0$ and $B_0 = B \cap \mathbb{M}_0$ in $\mathbb{M}_0$. In fact the vital part of this is actually how to deal with points in $A_0$ that being very close to points in $B_0$. Here is the detail:
First choose a r such that $U_0 = \cup D_r(b_i)$ has covered $B_0$. For every point $a$ in $A_0$, let $\delta_{a,n} = \frac {d_{a,n}^2}{4r}$, where $d_{a,n}$ is the distance between $a$ and $b_n$. So let 
$$\delta_a = \mathrm{min}\{ \frac {\mathrm{min}_n(\delta_{a,n}) }{2}, r \}$$
(to prevent the radium getting to $\infty$), $V_0=\{*\} \cup (\cup_{a\in A_0}D_{\delta_a}(a))$ is now by definition a open neighbourhood of $A_0$ with $V_0 \cap U_0 = \emptyset$.
The remaining is all about $\mathbb{M}_1$. I think from now on it's really "similar"  to the above case :) 
...Well, by definition all points in $A_0$ is isolated point for $B$, then for any $a$ in $A_0$ there exist a $r_a$ with $D_{r_a}(a)\cap B = \emptyset$. After fixing $\delta_a$ by $\mathrm{min}\{ \delta_a, r_a \}$, you can now choose $V_1$ and $U_1$ like before and $V = V_0 \cup V_1$,$U = U_0 \cup U_1$ are disjoint open neighbourhoods of $A$ and $B$. Strictly speaking, choose $U_1$ being an open neighbourhood of $B_1$ with $\overline U_1\cap \mathbb{M}_0 =\emptyset$ (this could be done since $\mathbb{M}_1$ is normal as a subspace of $\mathbb{R}^2$ and since $\mathbb{M}_1$ is a open neighbourhood of $B_1$, there exist a $V_1$ such that $B\subset V_1 \subset\overline V_1\subset \mathbb{M}_1$), then fix $\delta_a$ for every $a$ such that $D_{r_a}(a)\cap U_1 = \emptyset$.
