Moore plane / Niemytzki plane and the closed $G_\delta$ subspaces. I was working with the next exercise but really I'm so stuck. I have already some ideas but really, I don't know how to proced.

Let $\Gamma$ be the Moore plane. Prove that every closed set of $\Gamma$ is a $G_\delta$ set.

Firts, I think two cases. For the first one, let $X_0=\left\{(x,0):x\in\mathbb{R} \right\}$
1) If $A\subseteq \Gamma$ is a closed set and $A\cap X_0=\emptyset$ then $A$ is a closed usual set in the Euclidean topology (because the Euclidean topology is contained in the topology of the Moore plane) and this topology is metrizable. Thus, $A$ is a $G_\delta$ set. 
2) If $A\cap X_0\neq\emptyset$ then is equivalent to prove that $\Gamma\setminus A$ is a $F_\sigma$ set. But, then, we have that $(\Gamma\setminus A)\cap X_0=\emptyset$ and, again, $\Gamma\setminus A$ is an open set in the Euclidian topology, thus, $F_\sigma$ and therefore, $A$ is $G_\delta$
I don't know if that proof is correct (probably not).
The second one is:
If $A$ is an open set of $\Gamma$ then $A=H\cup J$ where $H\cap X_0=\emptyset$ and $H$ is an open usual set in the Euclidian topology and $J\subseteq X_0$ (is it true?). Then, every closed set $B$ is of the form $B=(F\cap T)\cup(F\cap (\Gamma\setminus X_0))$ (is it true?) where $F$ is an usual closed set in the Euclidean topology and $T\subseteq X_0$. We know that $(F\cap (\Gamma\setminus X_0))$ is $G_\delta$ because is a closed set in the Euclidean (metrizable) topology. Now, for a fixed $n\in\mathbb{N}$ consider $C_n=\left\{ \left\{ (z,0)\right\}\cup B_{\frac{1}{n}}\left(\left(z,\frac{1}{n} \right) \right):(z,0)\in T\cap F\right\}$. Therefore, $T\cap F=\displaystyle\bigcap_{n\in\mathbb{N}} C_n$ and thus $B$ is the union of two $G_\delta$ sets. We can conclude that $B$ is a $G_\delta$ set.
I really appreciate any help you can provide me.
 A: Let $X$ denote the Moore plane.

Let $L=\{(x,0)\in X\mid x\in \mathbb{R}\}$, and let $H=X\setminus L$.

Let $S \subseteq L$. 

The $X{\setminus}S$ is the union of $H$, together with all basic open disks of $X$ which are tangent to the $x$-axis at points of $L{\setminus}S$.

Thus, $X{\setminus}S$ is open in $X$, so $S$ is closed in $X$.

Therefore all subsets of $L$ are closed in $X$.

We want to show every closed subset of $X$ is $G_\delta$.

Equivalently, we'll show that every open subset of $X$ is $F_\sigma$.

Let $V$ be an open subset of $X$.

Then we can write $V=W\cup K$, where $W=V\cap H$, and $K=V\cap L$.

If $D\subset H$ is an open disk with center $p$, and radius $r$, then $D$ is the union of the countably many closed disks (closed both in $\mathbb{R^2}$ and in $X$) with center $p$, and rational radius less than $r$.

Thus, each such $D$ is $F_\sigma$.

Since $W$ is open in $H$, $W$ is also open in $\mathbb{R}^2$.

Thus, $W$ is the union of countably many open disks (open both in $\mathbb{R}^2$, and in $X$).

It follows that $W$ is $F_\sigma$.

Since $V = W\cup K$, and $K$ is closed in $X$, it follows that $V$ is $F_\sigma$, as was to be shown.

As regards your proof attempts, there are some problems . . .

You wrote:

1) If $A\subseteq \Gamma$ is a closed set and $A\cap X_0=\emptyset$ then $A$ is a closed usual set in the Euclidean topology (because the Euclidean topology is contained in the topology of the Moore plane) and this topology is metrizable. Thus, $A$ is a $G_\delta$ set.

You claim that since $A$ is closed in $\Gamma$, and $A\cap X_0=\emptyset$, it follows that $A$ is closed in the Euclidean topology. The claim is true, but not by the reasoning you gave. It's true that the Euclidean topology in the strict upper half plane $H$ is contained in the topology of the Moore plane, but that doesn't imply that a closed subset of $H$ in the Moore topology is closed in the Euclidean topology. Rather it implies that a closed subset of $H$ in the Euclidean topology is closed in the Moore topology.

You also wrote:

2) If $A\cap X_0\neq\emptyset$ then is equivalent to prove that $\Gamma\setminus A$ is a $F_\sigma$ set. But, then, we have that $(\Gamma\setminus A)\cap X_0=\emptyset$ and, again, $\Gamma\setminus A$ is an open set in the Euclidian topology, thus, $F_\sigma$ and therefore, $A$ is $G_\delta$

The error is your claim that $A\cap X_0\neq\emptyset$ implies $(\Gamma\setminus A)\cap X_0=\emptyset$. That's not a valid implication.

As for your "next idea", frankly, it's a little hard to follow.

In any case, my posted proof shows one way to resolve the problem.
