Separating closed sets in the bubble space. (a.k.a Moore plane a.ka. Niemytzki plane.) The bubble space generated by $B\subseteq\mathbb{R}$ (denoted $M(B)$) is defined as follows:

The base set of the space is the closed upper Euclidean plane minus the points not in $B$. Use the Euclidean topology on $\{(x,y)\,|\, y>0\}$, and neighborhoods of the points $x$ in $B$ are of the form $\{x\}\cup\{\text{open disk tangent to the axis at } x\}$ (like a bubble at $x$).

A $Q$-set is defined as follows: 

A $Q$-set is an uncountable subspace of $\mathbb{R}$ such that all of its subsets are $F_\sigma$.

(Just a fun fact: Under Martin's Axiom, every set of reals $X$ such that $\omega<|X|<2^{\omega}$ is a $Q$-set).
It is a famous result that $M(B)$ is a normal space when $B$ is a $Q$-set.
My problem: Assuming that for any $Y\subseteq B$ we can find disjoint open sets $U$ and $V$ with $Y\subseteq U$ and $B\smallsetminus Y\subseteq V$, it is apparently straightforward to finish the proof, i.e. to show that for arbitrary closed sets we can find the sets $U$ and $V$ separating them. This probably using the fact that $M(B)\smallsetminus B$ is just the Euclidean half plane in the subspace topology, and in particular normal.
I've been stuck for a while trying to show this. Can anybody offer some guidance?
 A: Suppose $B$ is a $Q$-set.
We begin by showing that for disjoint $X,Y\subseteq B$ there exist disjoint open sets in M$(B)$ separating them.
As $X$ and $Y$ are $F_\sigma$ relative to $B$ (as a subspace of $\mathbb{R}$) we may fix $\{X_i\}_{i<\omega}$ and $\{Y_i\}_{i<\omega}$ families of closed subsets of $\mathbb{R}$ such that $X=(\bigcup X_i)\cap B$ and $Y=(\bigcup Y_i)\cap B$.
For $n<\omega$ let $U_n$ be the union of the bubbles of radius $1$ at each $x\in X_n\cap B$.
We claim $\overline{U_n}\cap Y=\emptyset$.
Let $d(\cdot,\cdot)$ denote Euclidean distance.
To show the claim take $y\in Y$, notice that $d(y,X_n)>0$ as $X_n$ is closed, and let $x$ be a point in $X_n$ such that $d(y,x)=d(y,X_n).$
We can easily find a bubble at $y$ disjoint from the bubble of radius 1 at $x$ and this bubble will be disjoint from all of $U_n$, so the claim follows.
Likewise, fix $V_n$ such that $Y_n\subseteq V_n$ and $\overline{V_n}\cap X=\emptyset.$ 
For each $n$, we've found open sets $U_n$ and $V_n$ that contain $X_n$ and $Y_n$ respectively and satisfy $\overline{U_n}\cap Y =\emptyset = X\cap\overline{V_n}$.
Given this, a standard technique (used in, e.g., showing that regular Lindelöf spaces are normal) will give us a separation of $X$ and $Y$.
Consider the (open) sets
$$
\bigcup_{n<\omega}\Big(U_n\smallsetminus(\bigcup_{j≤n} \overline{V_j})\Big)
\text{ and }
\bigcup_{n<\omega}\Big(V_n\smallsetminus(\bigcup_{j≤n} \overline{U_j})\Big).
$$
If there were a point $p$ in the intersection then
$$
p\in U_n\smallsetminus(\bigcup_{j≤n} \overline{V_j}) \text{ and } p\in V_m\smallsetminus(\bigcup_{j≤m} \overline{U_j})
$$
for some $m$ and $n$, without loss of generality we may assume $n≤m$, and this gives $p\in U_n$ and $p\notin \overline{U_n}$.
Thus, this is the desired separation.
Now we treat the case of arbitrary disjoint closed sets.
Claim: Given $A\subseteq B$ and open $U$ containing $A$, there exists an open set containing $A$ such that its closure is contained in $U$.
By applying what we have just proven to $A$ and $B\smallsetminus A$, we obtain an open subset of $U$ containing $A$ such that its closure does not intersect $B\smallsetminus A$, and so, we may assume that $U$ is such a set.
Let $T(a,r)$ be the open disk of radius $r$ tangent to the $x$-axis at $a$.
There exists a set $\{r_a\mid a\in A\}\subseteq (0,1]$ such that
$$
E\stackrel{\text{def}}{=}\bigcup\{T(a,2r_a)\mid a\in A\}\subseteq U.
$$
Now, consider the set
$$
F =\bigcup\{T(a,r_a)\mid a\in A\}
$$
and suppose there is a point $q\in\overline{F}\smallsetminus E$.
We show $q$ must lie on the $x$-axis.
As this topology is finer than the Euclidean topology, $q$ is in the Euclidean closure of $F$.
Thus, there exists a sequence $(a_n)_n\subseteq A$ such that
$$
d(q,T(a_n,r_{a_n}))<\frac{1}{n}\quad\text{ and }\quad q\notin T(a_n,2r_{a_n}).
$$
As $(r_{a_n})_n\subseteq [0,1]$ and consequently $(a_n)_n$ is bounded, we may assume––by replacing $(a_n)_n$ with some subsequence––that they are convergent (in the usual Euclidean topology).
Let $\rho$ and $\alpha$ be their limit points.
The reader may convince him- or herself that 
$$
d(q,T(\alpha,\rho))=0\quad\text{ and }\quad q\notin T(\alpha,2\rho),
$$
which means $q$ is the tangency point of these disks, i.e., $q$ lies on the $x$-axis.
Now, the set $V=A\cup F$ is the open set we were looking for:
its closure is contained in the closure of $U$ so it does not intersect $B\smallsetminus A$, and its closure is contained in the closure of $F$ so a point in $\overline{V}\smallsetminus U$ would be in $\overline{F}\smallsetminus E$, and therefore on the $x$-axis, and hence in $A$.
This concludes the proof of the claim.
To continue, let $X,Y\subseteq$ M$(B)$ be arbitrary disjoint closed sets.
By the claim, there exist $U_1$ and $U_2$ open sets with disjoint closures separating $X\cap B$ and $Y\cap B$.
Then, the sets
$$
(X\cup \overline{U_1})\smallsetminus B \text{ and } (Y\cup\overline{U_2})\smallsetminus B 
$$
are disjoint closed subsets in the metric space $M(B)\smallsetminus B$, and as such, may be separated by disjoint open sets $V_1$ and $V_2$.
Thus, $V_1\cup U_1$ and $V_2\cup U_2$ form a separation of $X$ and $Y$.
