Corollary 2

Let $g$ be a Lie algebra, $\mathfrak{h}$ a subalgebra reductive in $g$ and $\mathfrak{k}$ a subalgebra of $\mathfrak{h}$ reductive in $\mathfrak{h}$. Then $\mathfrak{k}$ is reductive in $g$.

The representation $x \mapsto \mathrm{ad}_g x$ of $\mathfrak{h}$ on $g$ is semi-simple and hence its restriction to $\mathfrak{k}$ is semi-simple (Corollary 1).