Solution:
The assertion is false: we will show that the point $(\sqrt{2}, \sqrt{3})$ does not lie on a stable line.

Note that the slope of any stable line must be a rational number. Now assume for contradiction that $(a, b)$ lies on a stable line through $(\sqrt{2}, \sqrt{3})$, where $a$ and $b$ are both rational. Then $\frac{\sqrt{2}-a}{\sqrt{3}-b}=m$ for some rational number $m$, which leads us to
$$m b - a = m \sqrt{3} - \sqrt{2}.$$
Since $\sqrt{2}$ is not rational, we must have $m=0$. Then, squaring both sides gives
$$(m b - a)^2 = \left(3 m^2 + 2\right) - 2 m \sqrt{6}.$$
Since $m \neq 0$ this implies $\sqrt{6}$ is irrational, which is a contradiction.