Let $\mathfrak{g}$ be a finite-dimensional complex Lie algebra, $B$ be its killing form, and $$\ker(B)=\{X\in\mathfrak{g} | B(X,Y)=0,\forall Y\in \mathfrak{g}\}.$$
I want to show that $\ker(B)$ is a solvable ideal. First, we can easy check that $\ker(B)$ is an ideal of $\mathfrak{g}$ as it is a subalgebra which is closed under a bracket since $B(X,[Y,Z])=B([X,Y],Z)$ for all $X,Y,Z\in \mathfrak{g}$.
Next, I want to show that $\ker(B)$ is solvable. I procced the following way:
- We know that $\mathfrak{h}$ is solvable iff $B(\mathfrak{h},D\mathfrak{h})=0$ where $D\mathfrak{h}$ is the derived algebra of $\mathfrak{h}$ and $B$ is the killing form.
- By taking $\mathfrak{h}=\ker(B)$ and applying the previous statement, we are going to get that $\ker(B)$ is solvable.
Does it make sense?
Yes, the proof is correct. You are proving even more than is necessary because $B(\mathfrak{h},\mathfrak{h})=0$ if $\mathfrak{h}=\mathrm{Ker}(B)$.