 Solution

We prove Lie's theorem in the required form by induction on $\dim L$. If $L=0$, any nonzero vector works. Otherwise solvability gives $[L,L]\ne L$, so choose a codimension-one ideal $K\supseteq [L,L]$ and $x\in L\setminus K$. By induction there are $v\ne 0$ and a linear functional $\lambda :K\to \mathbb{C}$ with $av=\lambda (a)v$ for all $a\in K$.

Let $W=\mathrm{span}\{v,xv,x^{2}v,\dots \}$. Choose the cyclic basis $v,xv,\dots ,x^{r-1}v$ up to its first linear dependence. The commutator identity $a{x}^j=xa{x}^{j-1}+[a,x]x^{j-1}$, with $[a,x]\in K$, proves inductively thatHence $W$ is invariant under $K$ and $x$, and every $a\in K$ has trace $r\lambda (a)$ on $W$. In particular ${\mathrm{tr}}_W([x,a])=0$ implies $r\lambda ([x,a])=0$. Characteristic zero gives $\lambda ([x,a])=0$.

The simultaneous weight space $V_{\lambda }=\{w:aw=\lambda (a)w\forall a\in K\}$ is nonzero and stable under $x$, becauseOver $\mathbb{C}$, the restriction of $x$ to this nonzero finite-dimensional space has an eigenvector. That vector is an eigenvector of both $x$ and every $a\in K$, hence of every element of $L$.