This is a new version of a question I've asked before, but I've removed what I now know to be unnecessary details to make the question more readily accessible.
Let $P$ be the Euclidean plane above the $x$-axis (including the $x$-axis). For a point $a$ on the $x$-axis let $T(a,r)$="The open disk of radius $r$ tanget to the $x$-axis at $a$."
It seems to me (and I hope so) that this is true:
Let $b\in P$. If for every $\varepsilon>0$ there exists a point $a$ on the $x$-axis and a real number $r$ such that $$d(b,T(a,r))<\varepsilon\quad\text{ but }\quad b\notin T(a,2r),$$
then $b$ must lie on the $x$-axis.
(we may assume $r≤1$ but I don't think it will make a difference.)
It looks true as if $r$ is small the first condition forces $b$ near the $x$-axis, and if $r$ is big, the conjunction of both conditions forces $b$ near the $x$-axis as the only way to be near the inner disk but outside the outer disk is to be near the point of tangency. But I can't calculate the necessary formulas to make a proof.
Thank you for any help.
(Don't really know how to tag this. Included General-Topology as this is ultimately a question about the Niemytzki plane.)
You have this configuration below
By applying the Pythagorean theorem you can show that $$Y=\frac{\varepsilon^2}{2r}+\varepsilon$$ so for a big $r$ and small $\varepsilon$, point $b$ is constrained to be arbitrarily close to the $x$-axis.
Things get a little bit more interesting when $r$ is potentially small. The property you have let you pick a point $a$ and real $r$ for any $\varepsilon$, but if the $r$ value is very small compared to $\varepsilon^2$, then the bound $Y$ can become arbitrarily large. However in this configuration, this can never happen.
For $n\ge 1$, let $a_n$ and $r_n$ such that $d( b, T(a_n,r_n) )<\frac 1n$ and $b\notin T(a_n, 2r_n)$. Suppose that the sequence $(r_n)$ is bounded by $R$, then it admits a limit point $r$. Now every point of the sequence $(a_n)$ must lie in the closed disk of center $b$ and radius $R+1$, which is compact. So $(a_n)$ also admits a limit point $a$. You can then work out the limits and deduce $d( b,T(a,r) )=0$ and $b\notin T(a,2r)$. In other word $b$ must be the tangency point of those disks.