= Solution
We prove the common-zero-vector form of <Engel's theorem> by strong <induction> on the dimension of the acting <Lie algebra> of operators, allowing its action on a nonzero finite-dimensional space to have a kernel. The zero-dimensional case is immediate.
First, every proper <Lie subalgebra> $H<L$ has a strictly larger <normalizer of a Lie subalgebra>. The action of $H$ on $L/H$ by $h\cdot(x+H)=[h,x]+H$ consists of nilpotent operators by part (a). Its image has dimension at most $\dim H<\dim L$. By <induction> it has a nonzero common-zero <vector> $x+H$, giving $[H,x]\subseteq H$ with $x\notin H$. Hence $N_L(H)>H$.
Choose a maximal proper <Lie subalgebra> $H$. The preceding paragraph forces $N_L(H)=L$, so $H$ is an <ideal of a Lie algebra>. Moreover $L/H$ has dimension one: otherwise any nonzero one-dimensional subspace of $L/H$, which is a <Lie subalgebra>, would lift to an intermediate proper subalgebra. By <induction> the common kernel
$$
W=\{v\in V:Hv=0\}
$$
is nonzero. It is $L$-invariant, since $h(xv)=x(hv)+[h,x]v=0$ for $v\in W$, $h\in H$, $x\in L$. Take $x\notin H$. Its restriction to $W$ is nilpotent and so has a nonzero <vector> $v$ in its kernel. As $L=H+Kx$, this <vector> is killed by every element of $L$:
$$
\boxed{Lv=0\quad\text{for some }v\ne0.}
$$
This completes the <induction>. The underlying space is understood to be nonzero, as required for the asserted <vector> to exist.
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