= Solution
Use the nondegenerate restriction of the <Killing form> to $\mathfrak t$ to define $t_\alpha\in\mathfrak t$ by
$$
\kappa(t_\alpha,t)=\alpha(t)
\qquad(t\in\mathfrak t).
$$
Killing-form invariance and the <root-space decomposition> show that $\mathfrak g_\alpha$ pairs nondegenerately with $\mathfrak g_{-\alpha}$ and orthogonally with every other root space. Choose nonzero $e\in\mathfrak g_\alpha$ and $f\in\mathfrak g_{-\alpha}$ with $\kappa(e,f)\ne0$. For $t\in\mathfrak t$,
$$
\kappa([e,f],t)=\kappa(e,[f,t])
=\alpha(t)\kappa(e,f),
$$
so
$$
[e,f]=\kappa(e,f)t_\alpha\ne0.
$$
We need $\alpha(t_\alpha)\ne0$. If it were zero, the span of $e,f,t_\alpha$ would be a solvable Heisenberg-type Lie algebra with central commutator $[e,f]$. By <Lie theorem>, its adjoint action on $\mathfrak g$ can be upper triangularized, so $\operatorname{ad}[e,f]$ is nilpotent. But $[e,f]\in\mathfrak t$, and elements of the <Cartan subalgebra> act semisimply; hence $\operatorname{ad}[e,f]=0$. A semisimple Lie algebra has zero center, contradicting $[e,f]\ne0$.
Set
$$
h_\alpha=\frac{2t_\alpha}{\alpha(t_\alpha)}.
$$
Rescale $f$ so that $[e_\alpha,f_\alpha]=h_\alpha$. Since $e_\alpha$ and $f_\alpha$ lie in the $\alpha$ and $-\alpha$ root spaces,
$$
[h_\alpha,e_\alpha]=2e_\alpha,
\qquad
[h_\alpha,f_\alpha]=-2f_\alpha.
$$
Thus $\mathfrak m_\alpha=\langle e_\alpha,h_\alpha,f_\alpha\rangle$ is the <sl2 subalgebra associated with a root>.
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