 Diagonalizable adjoint operators force an algebraically closed Lie algebra to be abelian

Let $\mathfrak{g}$ be a finite-dimensional Lie algebra over an algebraically closed field. If every operator $\mathrm{ad}(x)$ of the Adjoint representation of a Lie algebra is diagonalizable, then $\mathfrak{g}$ is abelian. For if $[x,y]=\lambda y$ with $\lambda \ne 0$, then $\mathrm{ad}(y)x=-\lambda y\ne 0$ but $(\mathrm{ad}y{)}^{2}x=0$. This contradicts $\ker T^2=\ker T$ for a diagonalizable endomorphism $T$. Therefore every adjoint operator has only the eigenvalue zero and is diagonalizable, so every adjoint operator is zero. Hence every Lie bracket vanishes.