Why the radial topology on $ \mathbb R^2 $ is not second countable? 
A subset $ U\subset\mathbb R^2 $ is radially open if for every $ x\in U $ and every $ v\in\mathbb R^2 $, there exists $ \epsilon>0 $ such that $ x+sv\in U $ for every $ s\in (-\epsilon, \epsilon) $. Then the collection of radially open sets defines a topology on $ \mathbb R^2 $. Let $ X $ be $ \mathbb R^2 $ equipped with the radially open topology. 

Prove that $ X $ is not second countable.
The question is extracted from: (16) of https://math.uchicago.edu/~min/GRE/files/week2.pdf
 A: Make the observation that the circle is discrete. Justify that if $C$ is a circle, then for every $x\in C$, $(\mathbb{R}^2\setminus C)\cup\{x\}$ is open. Since this topology is clearly finer than the metric topology on $\mathbb{R}^2$, the circle is also closed. Since the circle is an uncountable closed discrete set, this topology is not Lindel$\ddot{\text{o}}$f and therefore not second countable either.
A: It is not even first-countable. Let $\{U_n:n\in\mathbb{N}\}$ be a countable family of neighbourhoods of $(0,0)$, say.
For $n\in\mathbb{N}$ let $v_n$ be the unit vector that makes an angle of $2^{-n}$ with the positive $x$-axis and let $\varepsilon_n$ be such that $\lambda v_n\in U_n$ whenever $|\lambda|<\varepsilon_n$.
Define $U$ to be the set that contains for every $n\in\mathbb{N}$ only the points $\lambda v_n$ with $|\lambda|<\varepsilon_n/2$ and for all other unitvectors, $v$, all points of the form $\lambda v$ with $\lambda>0$.
Show that $U$ is open in $X$ but that it contains none of the $U_n$.
