Indeed there's a much simpler way in characteristic zero.
Let $X$ be nilpotent. Clearly, this implies that $X\in [\mathfrak{g},\mathfrak{g}]$. Since it's enough to find one subgroup of $[G,G]$ doing the job, we can thus assume that $G$ is semisimple.
Now $\mathrm{ad}(X)$ being nilpotent, it preserves a complete flag (defined over $k$) in the linear space $\mathfrak{g}$. Let $H$ be the subgroup of $G$ preserving this flag (for the adjoint representation of $G$ on $\mathfrak{g}$), and acting as identity on all successive 1-dimensional quotients. Then $H$ is a $k$-defined unipotent subgroup of $G$, whose Lie algebra $\mathfrak{h}$ is the subalgebra of $\mathfrak{g}$ of those element preserving the flag, and acting (for the adjoint representation of $\mathfrak{g}$ on itself) as zero on all successive 1-dimensional quotients. Hence $X\in\mathfrak{h}$.