The Lie algebra of a semisimple group in characteristic zero is semisimple

If a nonzero solvable ideal $\mathfrak{r}$ exists, take its last nonzero derived term. Jacobi gives $[\mathfrak{g},[I,I]]\subseteq [I,I]$ for each ambient ideal $I$, so this term is an ambient ideal, and its next derived term being zero makes it commutative. Thus it suffices to exclude nonzero commutative ideals.

Let $\mathfrak{n}\subseteq \mathfrak{g}$ be a commutative ideal and put $\mathfrak{h}=\{x\in \mathfrak{g}:[x,\mathfrak{n}]=0\}$. Then $\mathfrak{n}\subseteq \mathfrak{h}$ because $\mathfrak{n}$ is commutative, and $\mathfrak{h}$ is an ideal of $\mathfrak{g}$: for $y\in \mathfrak{g}$, $x\in \mathfrak{h}$ and $n\in \mathfrak{n}$ one has $\left[\right[y,x],n]=[y,[x,n]]-[x,[y,n]]=0$ because $[x,n]=0$ and $[y,n]\in \mathfrak{n}$ is killed by $x$.

Choose a $k$-basis $y_1,\dots ,y_m$ of $\mathfrak{n}$ and let $H$ be the stabilizer of the point $(y_1,\dots ,y_m)$ of the rational representation ${\mathfrak{g}}^{\oplus m}$; this is a closed subgroup scheme of $G$, and $\mathrm{Lie}(H)=\mathfrak{h}$: a dual-number point $e^{\epsilon x}$ lies in $H$ exactly when $\mathrm{Ad}(e^{\epsilon x})y_i=y_i$ for all $i$, that is, by [F2], exactly when $[x,y_i]=0$ for all $i$.

Cartier's theorem [F4] makes the closed affine group $H$ smooth. Since $\mathfrak{h}=\mathrm{Lie}(H)$ is an ideal of $\mathfrak{g}$ by step 1.2, [F3] shows that $H^{\circ }$ is normal in $G$.

For any smooth connected affine $H_0$ in characteristic $0$, $Z(H_0)=\ker {\mathrm{Ad}}_{H_0}$. Indeed, over an algebraic closure a point of the latter has centralizer with full Lie algebra by [F4]. Cartier makes that centralizer smooth, so it has full dimension and equals connected $H_0$. Hence the geometric points of the adjoint kernel and centre agree; Cartier makes both subgroup schemes smooth and reduced, so they agree as schemes, and the equality descends to $k$. Differentiating the adjoint kernel now gives $\mathrm{Lie}Z(H_0)=\ker \mathrm{ad}=Z(\mathrm{Lie}{H}_0)$. Apply this to $H_0=H^{\circ }$: its centre is characteristic in $H^{\circ }$, hence normal in $G$ by step 3.1, and its Lie algebra is $Z(\mathfrak{h})$.

$Z(H^{\circ })$ is finite: its identity component is a connected commutative, hence solvable, normal subgroup of $G$, so it is contained in $R(G)=1$ by [F5]; since $Z(H^{\circ })$ is smooth of dimension $0$ in characteristic $0$, its Lie algebra is zero.

Finally $\mathfrak{n}\subseteq Z(\mathfrak{h})$: for $n\in \mathfrak{n}\subseteq \mathfrak{h}$ and $x\in \mathfrak{h}$ one has $[x,n]=0$ by the definition of $\mathfrak{h}$. Hence $\mathfrak{n}\subseteq Z(\mathfrak{h})=\mathrm{Lie}(Z(H^{\circ }))=0$, so $\mathfrak{n}=0$, and by step 1.1 the algebra $\mathfrak{g}$ has no nonzero solvable ideal.