= Solution
Every subalgebra of a <nilpotent Lie algebra> is nilpotent, since its lower central series is contained term by term in the ambient lower central series.
Conversely, if $L$ were not nilpotent, part (a) would give $x$ for which $\operatorname{ad}x$ is not nilpotent. Over an <algebraically closed field>, such an endomorphism has a nonzero <eigenvalue> $\lambda$ and <eigenvector> $y$. The vectors $x,y$ are independent because $[x,x]=0$ while $[x,y]=\lambda y\ne0$. Their span is a two-dimensional subalgebra, and its lower central series is $Ky$ at every stage after the first. It is therefore not nilpotent, a contradiction. Thus
$$
\boxed{L\text{ nilpotent}\iff\text{every two-dimensional subalgebra is nilpotent}.}
$$
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