 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 $\mathrm{ad}x$ 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 $[x,y]=\lambda y\ne 0$. 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