Corollary 1

Let $\mathfrak{q}$ be a maximal element of the set of subalgebras of $\mathfrak{g}$ distinct from $\mathfrak{g}$. Then $\mathfrak{q}$ is either parabolic or reductive in $\mathfrak{g}$.

We can assume that $\mathfrak{g}$ is a Lie subalgebra of $\mathfrak{g}\mathfrak{l}(V)$ for some finite dimensional vector space V. Let $\mathfrak{e}(\mathfrak{q})\subset \mathfrak{g}$ be the decomposable envelope of $\mathfrak{q}$. If $\mathfrak{e}(\mathfrak{q}) =\mathfrak{g},\mathfrak{q}$ is an ideal of $\mathfrak{g}$ (Chap. VII, §5, no. 2, Prop. 4), hence is semi-simple, and consequently $\mathfrak{q}$ is reductive in $\mathfrak{g}$. Assume that $\mathfrak{e}(\mathfrak{q})\not=\mathfrak{g}$. Then $\mathfrak{e}(\mathfrak{q}) =\mathfrak{q}$, so $\mathfrak{q}$ is decomposable. Assume that $\mathfrak{q}$ is not reductive in $\mathfrak{g}$. Let $\mathfrak{n}$ be the set of nilpotent elements of the radical of $\mathfrak{q}$. Then $\mathfrak{n}\not= 0$ (Chap. VII, §5, no. 3, Prop. 7 (i)). Let $\mathfrak{p}$ be the normalizer of $\mathfrak{n}$ in $\mathfrak{g}$. Then $\mathfrak{p}\supset \mathfrak{q}$, and $\mathfrak{p}\not=\mathfrak{g}$ since $\mathfrak{g}$ is semi-simple. Hence $\mathfrak{p}=\mathfrak{q}$. Thus $\mathfrak{q}$ is parabolic (Th. 1).