The nilradical is the set of all ad-nilpotent elements

Directly, ${\mathrm{ad}}_e(e)=0$, ${\mathrm{ad}}_e(f)=h$, and ${\mathrm{ad}}_e(h)=-2e$, so ${\mathrm{ad}}_e^3=0$. Likewise ${\mathrm{ad}}_f(e)=-h$, ${\mathrm{ad}}_f(f)=0$, and ${\mathrm{ad}}_f(h)=2f$, so ${\mathrm{ad}}_f^3=0$. Thus both $e$ and $f$ are ad-nilpotent.

Put $a=e+f$. Then $[a,h]=-2(e-f)$ and $[a,e-f]=-2h$, so $({\mathrm{ad}}_a{)}^2(h)=4h$. Since $h\mathrm{\ne }0$ and the field has characteristic zero, no power of ${\mathrm{ad}}_a$ is zero: its even powers send $h$ to $4^{r}h$. Hence $e+f$ is not ad-nilpotent.

The set of ad-nilpotent elements of ${\mathfrak{s}\mathfrak{l}}_2(k)$ contains $e$ and $f$ but not their sum, so it is not a linear subspace. By [L1] the nilradical is always a linear subspace, and therefore it cannot equal this set in the displayed example. This does not conflict with [L2], whose hypothesis quantifies over every element. The witness is finite and uses no choice.