Show that the radial interval topology without its origin is first countable but not second countable

Before we start, let me recall that the radial interval topology on $X=\mathbb{R}^2-\{(0,0)\}$ is the topology generated by intervals passing through the origin. For instance, the set $\{(t, t) \mid 1<t<2\}$ is an open set in this topology.

As you said, one way to justify the fact that $X$ is first-countable is to see each ray individually. More precisely, given $x \in X$, one can show that the map $$ f\colon (0, \infty) \to X, \qquad t \mapsto tx $$ is an open embedding. In fact, we can see that the map is injective, continuous and open (by definition of the topology). Here is really important that we are avoiding zero, as then is not clear that the map is continuous. As $(0, \infty)$ is first-countable, the same holds for its image in $X$; in particular, the first-countable axiom holds for $x \in X$. As $x \in X$ is arbitrary, this proves the result.

Now, to show that $X$ is not second countable, I think there are two possible ways to show it. The first one is to show that the natural map $S^1 \times (0, \infty) \to X, (v, t) \mapsto tv$ is a homemorphism, if $S^1$ carries the discrete topology (instead of the Euclidean one), while $(0, \infty)$ carries the Euclidean topology. I'll leave this one as an exercise; then you will have to show that if a product of non-empty topological spaces is second countable, then each one has to be second countable as well.

The second proof I have in mind is to show directly that there is no countable base. Assume by contradiction that such base exists, say $\{U_i\}$. The idea is to produce an open set $U$ that does not contain any of the $U_i$. For this, assume first that each $U_i$ has infintely many "rays". We will use a "diagonal argument", as follows:

• For $U_1$, choose some $\theta_1$ such that $U_1$ contains a ray in the $\theta_1$ angle. Fix an open interval $I_1$ small enough, such that $I_1$ is smaller than the component on the $\theta_1$ direction. (For instance, if $\theta=45^\circ$ and the component of $U_1$ is $\{(t, t) \mid 1<t<2, 3<t<4\}$, we take $\{(t, t) \mid 1<t<1.5\}$.)
• For $U_2$, choose some $\theta_2\neq \theta_1$ such that $U_1$ contains a ray in the $\theta_2$ angle. As before, fix some interval $I_2$ small enough.
• In general, we choose some $\theta_i \neq \theta_1, \dots, \theta_{i-1}$, and an interval $I_i$ in that direction small enough to not contain the corresponding component of $U_i$. (This uses that each $U_i$ has infinitely many rays!)

This way, let $U=\bigcup_i I_i$. Note that $U$ does not contain $U_i$, as in the $\theta_i$ direction, the set $U$ is smaller. This shows that $X$ does not have a countable basis.

Try to (slightly) modify the argument to include basis where some elements have finitely many rays!

Take the set of all open rays starting at the origin. It's an uncountable set of disjoint open subsets of your space. So, it's an open cover whose only subcover is itself. In particular, your space is not Lindelöf, and therefore it is not second countable.

But it is first countable. Given $p\in\Bbb R^2\setminus\{(0,0)\}$, consider all line segments of the form$$B_{1/n}(p)\cap\{\lambda p\mid\lambda>0\}\quad\text{with }n\in\Bbb N\text{ such that }\frac1n<\|p\|.$$It is a countable fundamental system of neighborhoods of $p$.