The Bohr-Van Leeuwen theorem I don't think is relevant here since it is a thermodynamic statement and really $\mu$ is conserved for individual particles .
The derivation I would do is specific to the case of converging magnetic fields, such as in a magnetic mirror.
Assume the particle is gyrating with velocity $v_L$ around some $B_\parallel$, additionally with some parallel velocity $v_\parallel$.
The $B$ does no work, so
$$\frac{1}{2}m\frac{d}{dt}(v_\perp^2 + v_\parallel^2)=0$$
Then we simplify $\frac{d}{dt}(v_\parallel^2)=2v_\parallel \frac{d}{dt}v_\parallel$.
The $B$ is divergence free, so roughly one must have a $B_\perp$ pointing inwards, $B_\perp=\frac{r}{2}\frac{d}{dz}B_\parallel$.
The Lorentz force averages to
$$F_\parallel=\frac{d}{dt}v_\parallel=-\frac{e v_\perp B_\perp}{m}$$
$$\frac{d}{dt}v_\parallel=-\frac{e v_\perp r}{2m}\frac{d}{dz}B_\parallel$$
Multiplying by $v_\parallel$ (the velocity along the z axis),
$$v_\parallel \frac{d}{dt}v_\parallel=-\frac{e v_\perp r}{2m}v_\parallel\frac{d}{dz}B_\parallel$$
Or
$$2v_\parallel \frac{d}{dt}v_\parallel=-\frac{e v_\perp r}{m}\frac{d}{dt}B_\parallel$$
We can insert this into our energy equation. Recall $r=v_\perp/\Omega$ where $\Omega$ is the gyration frequency $eB/m$. Simplifying gives
$$0=\frac{d}{dt}v_\perp^2 -\frac{ v^2_\perp }{ B_\parallel}\frac{d}{dt}B_\parallel,$$
$$0=B_\parallel\frac{d}{dt}\frac{v_\perp^2}{B_\parallel}. $$
Inserting constants gives
$$0=\frac{d}{dt}\frac{1/2 m v_\perp^2}{B_\parallel}=\frac{d}{dt}\mu.$$
