The radical is characteristic and its quotient has zero radical

If $f$ is an automorphism of $\mathfrak{g}$, then $f(\mathfrak{r})$ is an ideal and its derived series is the image of the derived series of $\mathfrak{r}$, so it is solvable. Maximality in [L1], justified by [L2], gives $f(\mathfrak{r})\subseteq \mathfrak{r}$. Applying the same argument to $f^{-1}$ gives the reverse inclusion, hence equality.

Let $\bar{\mathfrak{a}}$ be a solvable ideal of $\mathfrak{g}\mathrm{/}\mathfrak{r}$, and let $\mathfrak{a}$ be its inverse image under the quotient map [L4]. Then $\mathfrak{a}$ is an ideal containing $\mathfrak{r}$ and $\mathfrak{a}\mathrm{/}\mathfrak{r}=\bar{\mathfrak{a}}$ is solvable. Since $\mathfrak{r}$ is solvable, extension closure [L3] makes $\mathfrak{a}$ solvable. By [L1], $\mathfrak{a}\subseteq \mathfrak{r}$, so equality holds and $\bar{\mathfrak{a}}=0$. Thus the quotient's largest solvable ideal is zero. This includes $\mathfrak{r}=0$ and $\mathfrak{r}=\mathfrak{g}$.