 Solution

If $\mathfrak{g}$ is nilpotent, every Lie subalgebra $\mathfrak{a}$ is nilpotent, since ${\mathfrak{a}}^j\subseteq {\mathfrak{g}}^j$ in the respective lower central series of a Lie algebra. A two-dimensional nilpotent algebra has nonzero centre by part (i). If $0\ne z$ is central, complete it to a basis $(z,u)$; every bracket of these basis elements is zero. Thus $\mathfrak{a}$ is abelian.

Conversely, suppose $\mathfrak{g}$ is not nilpotent. By Engel's theorem, some $\mathrm{ad}(x)$ is not nilpotent. Since the field is algebraically closed, it has an eigenvalue $\lambda \ne 0$ with eigenvector $y\ne 0$. ThusThe vectors $x,y$ are independent: otherwise their bracket would be zero. Their span is therefore a two-dimensional Lie subalgebra and it is nonabelian. This contradicts the hypothesis. Hence