Why is curl of current density $\nabla \times \vec{J}$ equal zero?

I think the better way to derive this is to first observe the Biot-Savart law, $$ \mathbf B(\mathbf r)=\frac{\mu_0}{4\pi}\int\mathbf J(\mathbf r')\times\frac{\hat{r}}{r^2}\,\mathrm dV'\tag{1} $$ Since $$ \frac{\hat r}{r^2}=-\nabla_r\left(\frac1r\right) $$ (your text may derive this, if not you can prove it by starting with the RHS), we can write (1) as $$ \mathbf B(\mathbf r)=-\frac{\mu_0}{4\pi}\int\mathbf J(\mathbf r')\times\nabla_r\left(\frac{1}{r}\right)\,\mathrm dV'\tag{2} $$ Since $\mathbf J$ is a function or $r'$ and not $r$, we can put it inside the parenthesis and swap the order of the cross product (i.e., $\mathbf J\times\nabla=-\nabla\times\mathbf J$), $$ \mathbf B(\mathbf r)=\frac{\mu_0}{4\pi}\int\nabla_r\times\frac{\mathbf J(r')}{r}\,\mathrm dV'\tag{3}=\nabla_r\times\frac{\mu_0}{4\pi}\int\frac{\mathbf J(r')}{r}\,\mathrm dV' $$ Then we can define the vector potential as $$ \mathbf A(\mathbf r)=\frac{\mu_0}{4\pi}\int\frac{\mathbf J(r')}{r}\,\mathrm dV' $$ To get $$ \mathbf B(\mathbf r)=\nabla\times\mathbf A(\mathbf r)\tag{4} $$ where we drop the subscript $r$ because it's implied that it's over $\mathbf r$.

That proof over, we can take the divergence of (4): $$ \nabla\cdot\mathbf B=\nabla\cdot\nabla\times\mathbf A\equiv0 $$ by the fact that the divergence of every curl is identically zero (worth the effort to prove this).

I also dislike when authors claim things to be obvious. If it's so simple, then why not just write it out.

Anyhow, regarding this specific case. If you go to the definition of the curl you will see that this is a collection of partial derivatives with respect to position.

So to claim that the curl is zero is to claim that the velocity is independent of the particles position, ie. it is assumed that there are no other fields present, be it gravitaional or electrical.