Show equivalence classes are the orbits of a group action 
If we define the projective plane $\mathbb{R}\mathbb{P}^2$ as the set
of all straight lines through the origin of $\mathbb{R}^3$, then
there's a surjective map
$\pi: \mathbb{R}^3 \backslash \{0\} \to \mathbb{R}\mathbb{P}^2, x \mapsto l_x$,
where $l_x$ is the line through the origin, and the point $x$, such
that $l_x = \{\lambda x : \lambda \in \mathbb{R}\} \subset \mathbb{R}^3$. And $\pi$ induces a quotient topology on
$\mathbb{R}\mathbb{P}^2$.

How does one prove that the equivalence classes of the equivalence relation "$x \sim y$ iff $\pi(x) = \pi(y)$" are the orbits of a group action of the multiplicative group $\mathbb{R}^*$ (set of real numbers without 0) on $\mathbb{R}\backslash \{0\}$. I have no idea how to approach it, my algebra is very rusty, and I'm struggling with finding the intuition for quotient topologies.
Am I right in thinking the equivalence classes are homeomorphic with the orbits? Or do I need to show that for every $x$ the set $l_x$ is equal to the set of the orbits: $x\mathbb{R}^* = \{ gx : g \in \mathbb{R}^* \}$? Are the equivalence classes the lines?
 A: For any $x_0 \in \Bbb{R}^3 \setminus \{0\}$, its equivalence class is
\begin{align*}
[x_0]_\sim 
&= \{x \in \Bbb{R}^3 \setminus \{0\} : x \sim x_0\} \\
&= \{x \in \Bbb{R}^3 \setminus \{0\} : l_x = l_{x_0}\} \\
&= \{x \in \Bbb{R}^3 \setminus \{0\} : (\exists \lambda \in \Bbb{R}^*) \ x = \lambda x_0\} \tag{1} \\
&= \{\lambda x_0 : \lambda \in \Bbb{R}^*\}.
\end{align*}
Let’s prove the equality $(1)$. To do this, let $x \in \Bbb{R}^3 \setminus \{0\}$.

*

*If $l_x = l_{x_0}$, then $x \in l_{x_0}$, so there is $\lambda \in \Bbb{R}$ such that $x = \lambda x_0$.
Note that $\lambda \neq 0$ since $x \neq 0$.
This proves the inclusion ‘$\subseteq$’.

*On the other hand, if $x = \lambda x_0$ for some $\lambda \in \Bbb{R}^*$, then for each $t \in \Bbb{R}$ one has that $tx = t \lambda x_0 \in l_{x_0}$ (so $l_x \subseteq l_{x_0}$) and that $tx_0 = t\lambda^{-1}x \in l_x$ (so $l_{x_0} \subseteq l_x$), hence $l_x = l_{x_0}$.
This proves ‘$\supseteq$’.

Now, $\Bbb{R}^*$ acts on $\Bbb{R}^3 \setminus \{0\}$ by left-multiplication, and the orbit of $x_0$ is (by definition) the set $\Bbb{R}^*x_0 = \{\lambda x_0 : \lambda \in \Bbb{R}^*\}$.
