 Solution

Suppose $\mathrm{ad}(x)$ has a nonzero eigenvalue $\lambda$ and corresponding vector $y$. Then $[x,y]=\lambda y$, soFor a diagonalizable endomorphism $T$, its eigenspace decomposition shows $\ker T^2=\ker T$. The displayed equations therefore contradict diagonalizability of $\mathrm{ad}(y)$.

Thus no $\mathrm{ad}(x)$ has a nonzero eigenvalue. Each is diagonalizable over the algebraically closed field, so each is the zero endomorphism. Hence $[x,z]=0$ for every $x,z\in \mathfrak{g}$, proving