Why is Niemytzki plane locally metrizable? The answer https://math.stackexchange.com/a/4411293/32337 includes the assertion that the Niemytzki plane $\Gamma$ (also known as the Moore plane, and as the tangent disk space) is locally metrizable.
What is a metric on a basic open neighborhood $N = \{(x, 0\} \cup B_{r}((x,0))$ of a point $(x, 0)$ on the horizontal axis that is compatible with the topology of $\Gamma$?  [Here $B_{r}((x,0))$ denotes the usual Euclidean open ball of radius $r$ at the indicated point.]
 A: The Moore plane is Hausdorff and completely regular, and therefore $N$ is also Hausdorff and completely regular. In particular, $N$ is Hausdorff and regular. Besides, it is clearly second countable. But then, by Urysohn's metrization theorem, $N$ is metrizable.
A: Any point $(x',y')$ in $N$, except $(x,0)$, lies on the boundary of some $B_\epsilon((x,\epsilon))$.  Define $\rho(x',y')$ to be that $\epsilon$, and $\rho(x,0) = 0$. The map $(x',y') \mapsto (x',y', \rho (x',y'))$ embeds $N$ in $\mathbb R^3$ (as a skewed cone with a slit down the side).  Transferring the standard metric back gives this metric on $N$:
$d((x_1,y_1),(x_2,y_2)) = \sqrt {(x_2-x_1)^2 + (y_2-y_1)^2 +(\rho(x_2,y_2) - \rho(x_1,y_1))^2}$
ETA further explanation.
Let $(x,\epsilon)$ be the point where the perpendicular bisector of $(x',y')$ and $(x,0)$ crosses the line through $(x,0)$ parallel to the y-axis. Then $(x',y')$ is on the boundary of $B_\epsilon((x,\epsilon))$. Simple geometry shows that $\epsilon = [(x'-x)^2 + y'^2]/2y'$.
Let's call the inclusion map $\phi$ and its image $M$. $\phi^{-1}$ is the restriction to $M$ of the projection onto the (x,y)-plane. $N \setminus \{(x,0)\}$ and $M\setminus\{(x,0,0)\}$ have the standard topology so $\phi^{-1}|_{M\setminus\{(x,0,0)\}}$ is continuous.Its restriction to any compact subset of $M\setminus\{(x,0,0)\}$ will be closed which is enough to show it is a homeomorphism between $M\setminus\{(x,0,0)\}$ and $N\setminus\{(x,0)\}$.
That leaves only the behaviour at $(x,0)$ to be considered. Let $C_\epsilon$ be the intersection of the $\epsilon$-ball in $\mathbb R^3$ centered on $(x,0,0)$ with $M$. It is a cone-like shape which lies below the plane $z=\epsilon$ and contains all of that part of $M$ which is below $z=\epsilon/\sqrt{5}$. Hence
$B_{\epsilon/\sqrt{5}}((x,\epsilon/\sqrt{5})) \cup \{(x,0)\} \subset \phi^{-1}(C_\epsilon) \subset B_\epsilon((x,\epsilon)) \cup \{(x,0)\}$.
That shows that $\phi$ and $\phi^{-1}$ are continuous at $(x,0)$ and $(x,0,0)$.

Added later:
Here's an alternative metric which is perhaps simpler to visualise. It has the virtue that the basic tangent disc nbhds. of $(x,0)$ are the open balls around it in this metric.
Let $d_1((x_1,y_1),(x_2,y_2)) = \max(|x_2-x_1|,\frac 12 |y_2-y_1|,|\rho(x_2,y_2) - \rho(x_1,y_1)|)$.
The balls for $d_1$ around points in $N \setminus \{(x,0)\}$ are intersections of the crescent-shaped areas between tangent discs with open rectangles, giving the usual topology; around $(x,0)$ they are the whole tangent discs (with $(x,0)$ attached). So overall $d_1$ induces the required topology on $N$.
Yet another approach is to show that $N$ can be embedded in $\mathbb R^2$ as an open disc with a slit, with the Euclidean topology (i.e with the points on one radius removed but the centre point retained). The map $N \rightarrow \mathbb R^2$ given by
$(x',y') \mapsto (x',y'-\rho(x',y'))$
does the trick. The metric induced on $N$ is
$d_2((x_1,y_1),(x_2,y_2)) = \sqrt{(x_2-x_1)^2 +(y_2-\rho(x_2,y_2) -y_1 +\rho(x_1,y_1))^2}$.
A more complicated formula, but probably the easiest way to visualise $N$.
