A nilpotent acting basis suffices for Engel's theorem

Put $u=h+e-f$. The three vectors $e,f,u$ form a basis: comparison of the $h$-coefficient in $ae+bf+cu=0$ first gives $c=0$, and then $a=b=0$. Directly, ${\mathrm{ad}}_e(e)=0$, ${\mathrm{ad}}_e(f)=h$, and ${\mathrm{ad}}_e(h)=-2e$, so ${\mathrm{ad}}_e^3=0$; similarly ${\mathrm{ad}}_f(e)=-h$, ${\mathrm{ad}}_f(f)=0$, and ${\mathrm{ad}}_f(h)=2f$, so ${\mathrm{ad}}_f^3=0$.

The remaining brackets are $[u,e]=2e+h$, $[u,f]=h-2f$, and $[u,h]=-2e-2f$. Applying ${\mathrm{ad}}_u$ once more sends these three values respectively to $2u,-2u,-4u$, and $[u,u]=0$; hence ${\mathrm{ad}}_u^3=0$. Thus every member of the chosen basis $e,f,u$ acts nilpotently.

Yet $h=u-e+f$ belongs to their span and ${\mathrm{ad}}_h(e)=2e$, so $({\mathrm{ad}}_h{)}^r(e)=2^{r}e\mathrm{\ne }0$ for every $r\ge 1$ in characteristic zero. Therefore ${\mathrm{ad}}_h$ is not nilpotent, the adjoint representation is not nil in the sense of [L2], and $\mathfrak{g}$ is not nilpotent by [L1]. This basis is the required counterexample; all calculations are finite and choice-free.