The commutator with the radical lies in the nilradical

First let $\mathfrak{s}$ be any finite-dimensional solvable characteristic-zero Lie algebra and $D\in \mathrm{Der}(\mathfrak{s})$. Form $\mathfrak{h}=kt{⋉}_D\mathfrak{s}$ as in [L3], where $t$ acts by $D$. The ideal $\mathfrak{s}$ is solvable and $\mathfrak{h}\mathrm{/}\mathfrak{s}\cong kt$ is abelian, so [L4] makes $\mathfrak{h}$ solvable. Its derived algebra ${\mathfrak{h}}^{\mathrm{\prime }}$ is a nilpotent ideal by [L5], and ${\mathfrak{h}}^{\mathrm{\prime }}\subseteq \mathfrak{s}$ because the quotient is abelian. Hence ${\mathfrak{h}}^{\mathrm{\prime }}$ is a nilpotent ideal of $\mathfrak{s}$, so [L2] gives ${\mathfrak{h}}^{\mathrm{\prime }}\subseteq \mathrm{nilrad}(\mathfrak{s})$. Since $D(y)=[t,y]\in {\mathfrak{h}}^{\mathrm{\prime }}$ for every $y\in \mathfrak{s}$, we have $D(\mathfrak{s})\subseteq \mathrm{nilrad}(\mathfrak{s})$.

Put $\mathfrak{r}=\mathrm{rad}(\mathfrak{g})$ and ${\mathfrak{n}}_r=\mathrm{nilrad}(\mathfrak{r})$. For each $x\in \mathfrak{g}$, ideality of $\mathfrak{r}$ makes ${\mathrm{ad}}_x{\mathrm{\mid }}_{\mathfrak{r}}$ a derivation of the solvable Lie algebra $\mathfrak{r}$. Step 1.1 therefore gives $[x,\mathfrak{r}]\subseteq {\mathfrak{n}}_r$, and in particular $[x,{\mathfrak{n}}_r]\subseteq {\mathfrak{n}}_r$. Thus ${\mathfrak{n}}_r$ is a nilpotent ideal of the ambient algebra $\mathfrak{g}$.

By maximality in [L2], step 2.1 gives ${\mathfrak{n}}_r\subseteq \mathrm{nilrad}(\mathfrak{g})$. Conversely, $\mathrm{nilrad}(\mathfrak{g})$ is solvable by [L6], hence lies in $\mathfrak{r}$ by [L1]; inside $\mathfrak{r}$ it is still a nilpotent ideal, so maximality gives $\mathrm{nilrad}(\mathfrak{g})\subseteq {\mathfrak{n}}_r$. Therefore the two nilradicals are equal. Combining this equality with $[\mathfrak{g},\mathfrak{r}]\subseteq {\mathfrak{n}}_r$ from step 2.1 proves the claim. If $\mathfrak{r}=0$ or $\mathfrak{g}=0$, every subspace displayed here is zero. No choice principle is used.