This is obvious since there is a more general result for semi-simple Lie algebras, but I would like to prove it directly.
So suppose that the Killing form $\kappa$ is degenerate and let $X\in L$ be such that $\kappa(X,Y)=0$ for all $Y \in L$.
I would now like to use $X$ to construct a non-trivial ideal of $L$, thereby showing that it is not simple.
The ideal $\mathrm{span}_{\mathbb{F}}\{[X,Y] : Y \in L\} $ seems promising, but I haven't been able to show that it is non-trivial. I'm not sure if I am on the right track.
UPDATE Dietrich Burde's answer shows that this is in fact not true if for example the field $\mathbb{F}$ has characteristic 3. So in this case I assume I am only considering fields of characteristic 0, and moreover one may assume that $\kappa$ does not vanish identically.
Best Answer
Suppose that the characteristic is zero. $I=\{X:K(X,.)=0\}$ is an ideal. To see this, note that $K([X,Y],Z)=K(X,[Y,Z])$. If $K(X,.)=0,$ for every $Y,Z$, $K([X,Y],Z)=K(X,[Y,Z])=0$. This implies that $[X,Y]\in I$. Suppose that $I$ is not $\{0\}$. Then, since the Lie algebra ${\cal G}$ is simple, $I={\cal G}$ and $K=0$. This implies that ${\cal G}$ is solvable by the Cartan criterion. Contradiction.
https://en.wikipedia.org/wiki/Cartan%27s_criterion#Cartan.27s_criterion_for_solvability