The adjoint highest weight is the highest root

Since $\mathfrak{g}$ is nonabelian, its derived ideal $[\mathfrak{g},\mathfrak{g}]$ is nonzero. Simplicity and [L5] give $[\mathfrak{g},\mathfrak{g}]=\mathfrak{g}$, so every term of the derived series equals $\mathfrak{g}\mathrm{\ne }0$. Thus $\mathfrak{g}$ is not solvable. Its radical, being an ideal, is either zero or $\mathfrak{g}$; the latter would make $\mathfrak{g}$ solvable. Hence the radical is zero and $\mathfrak{g}$ is semisimple, licensing the semisimple root-space interfaces [L3].

By [L1] subrepresentations of the adjoint module are precisely ideals. Simplicity and nonzeroness imply this representation is irreducible.

Apply [L3] using step 1.1. The weights of the adjoint module are the roots on their root spaces and zero on $\mathfrak{h}$. In particular the specified root $\theta$ has a nonzero one-dimensional weight space. Choose $0\mathrm{\ne }x\in {\mathfrak{g}}_{\theta }$. For every positive root $\alpha$, the bracket $[{\mathfrak{g}}_{\alpha },x]$ lies in ${\mathfrak{g}}_{\alpha +\theta }$. Since $\alpha +\theta$ is nonzero and strictly greater than $\theta$ in the root order, it cannot be a root by maximality, so this bracket vanishes. Therefore ${\mathfrak{n}}^{+}x=0$ and $Hx=\theta (H)x$ for every $H\in \mathfrak{h}$.

The subrepresentation generated by the nonzero $x$ of step 2.1 is nonzero, hence is the entire adjoint representation by step 1.2. Thus $x$ is a highest weight vector of weight $\theta$ generating the module, exactly the definition of a highest weight module (Highest-weight vectors and modules). The proof uses the given maximal root directly and does not presume that an arbitrary irreducible module has a unique maximal weight. Simplicity excludes both the zero algebra and a one-dimensional abelian algebra; no additional choice beyond [A1] is needed to select one nonzero vector in the given root line.