Consider the map $\mathbb{R}\mapsto\mathbb{R}^2,g\mapsto g.x$, where $x$ is any point on the figure eight. The image of this map is orbit of $x$, i.e. the figure eight. Consider the stabilisator of $x$, which is a subgroup of $\mathbb{R}$, so it is either dense or a multiple of $\mathbb{Z}$. It can't be dense, for otherwise the action would be trivial by continuity, but the orbit isn't a singleton.
If the stabilistaor is a non-zero multiple of $\mathbb{Z}$, the map descends to a continuous bijection from $\mathbb{R}/r\mathbb{Z}$ to the figure eight. But the quotient is compact, so this is a…
Another trick or threat.
$$
2\ln(x-a) = \ln(x-b) + \ln(x+b)
$$
Applying $\frac{\mathrm{d}}{\mathrm{d}x}$ to both sides, we have
\begin{align}
\frac{2}{x-a} &= \frac{1}{x-b} + \frac{1}{x+b}\\
\frac{2}{x-a} &= \frac{2x}{x^2-b^2}\\
x^2-b^2 &= x^2-ax\\
a &=\frac{b^2}{x}
\end{align}
