= Solution
For a complex <semisimple Lie algebra> $\mathfrak g$, a <Cartan subalgebra> $\mathfrak h$ is a maximal abelian subalgebra consisting of elements whose <Adjoint representation> matrices are diagonalizable. Equivalently it is a nilpotent self-normalizing <Lie subalgebra>. The <root-space decomposition> is
$$
\mathfrak g=\mathfrak h\oplus\bigoplus_{\alpha\in R}\mathfrak g_\alpha,\qquad\mathfrak g_\alpha=\{x:[h,x]=\alpha(h)x\text{ for every }h\in\mathfrak h\},
$$
where the roots are the nonzero weights of that <Adjoint representation>.
We use these properties of the <Killing form> $B$: it is nondegenerate on $\mathfrak g$ and on $\mathfrak h$, it is invariant, distinct <root spaces> are orthogonal unless their roots sum to zero, and $B$ pairs $\mathfrak g_\alpha$ nondegenerately with $\mathfrak g_{-\alpha}$. Define $t_\alpha$ by $B(t_\alpha,h)=\alpha(h)$, and choose $e\in\mathfrak g_\alpha$, $f\in\mathfrak g_{-\alpha}$ with $B(e,f)=1$. Since weights add, $[e,f]\in\mathfrak h$, and invariance gives
$$
B([e,f],h)=B(e,[f,h])=\alpha(h),\qquad[e,f]=t_\alpha.
$$
The <nonisotropic root lemma> shows $\alpha(t_\alpha)\ne0$. Here is its short proof: if this number were zero, the span of $e,f,t_\alpha$ would be a <solvable Lie algebra> with $t_\alpha$ central. Apply the <Lie theorem> to its action on $\mathfrak g$. The commutator $\operatorname{ad}t_\alpha=[\operatorname{ad}e,\operatorname{ad}f]$ would be strictly upper triangular and hence nilpotent. But $t_\alpha\in\mathfrak h$ makes it diagonalizable. It would therefore be zero, putting $t_\alpha$ in the zero <center of a Lie algebra> of $\mathfrak g$, contrary to its definition. Set
$$
H_\alpha=\frac{2t_\alpha}{\alpha(t_\alpha)},\qquad X_\alpha=e,\qquad Y_\alpha=\frac{2f}{\alpha(t_\alpha)}.
$$
Then
$$
\boxed{[H_\alpha,X_\alpha]=2X_\alpha,\quad[H_\alpha,Y_\alpha]=-2Y_\alpha,\quad[X_\alpha,Y_\alpha]=H_\alpha,}
$$
so their span is the <sl2 subalgebra associated with a root>.
The abstract reduced <crystallographic root system> axioms are as follows. In a finite-dimensional real <inner product space> $E$, the set $R$ is finite, consists of nonzero vectors and spans $E$; for every $\alpha\in R$, one has $R\cap\mathbb R\alpha=\{\alpha,-\alpha\}$; the <root reflection> $s_\alpha(\beta)=\beta-2(\beta,\alpha)\alpha/(\alpha,\alpha)$ preserves $R$; and $2(\beta,\alpha)/(\alpha,\alpha)$ is an <integer> for every pair of roots. We verify the <positive-definite> real form as well as these axioms, rather than assuming the complex <Killing form> is already positive.
First $R$ is finite and has no zero element by its definition. The roots span $\mathfrak h^*$ over $\mathbb C$: any $h\in\mathfrak h$ annihilated by all roots commutes with the whole <root-space decomposition> and is central, hence zero. Therefore the $t_\alpha$, and also the $H_\alpha$, span $\mathfrak h$ over $\mathbb C$. For every root $\beta$, the <classification of finite-dimensional sl2 representations> applied to the <Adjoint representation> of the root subalgebra gives $\beta(H_\alpha)\in\mathbb Z$. Put $\mathfrak h_{\mathbb R}=\operatorname{span}_{\mathbb R}\{H_\alpha:\alpha\in R\}$. Every root takes real values on this space, and
$$
B(h,h)=\operatorname{tr}((\operatorname{ad}h)^2)=\sum_{\beta\in R}(\dim\mathfrak g_\beta)\beta(h)^2>0\qquad(0\ne h\in\mathfrak h_{\mathbb R}).
$$
The strict inequality follows because the roots span $\mathfrak h^*$. This also shows that the complexification of $\mathfrak h_{\mathbb R}$ injects into $\mathfrak h$: an equality $h_1+ih_2=0$ with real $h_j$ would contradict positivity of $B(h_1,h_1)$ and $B(h_2,h_2)$. Since its complex span is all of $\mathfrak h$, it is a real form. Moreover $B(H_\alpha,H_\alpha)=4/\alpha(t_\alpha)>0$, so $t_\alpha$ is a real scalar multiple of $H_\alpha$. The dual <inner product> thus makes $E=\operatorname{span}_{\mathbb R}R$ a Euclidean space, as in the <Euclidean subspace of a Cartan subalgebra>.
To prove reducedness without assuming it, consider the root-subalgebra module
$$
M_\alpha=\mathfrak h\oplus\bigoplus_{j\in\mathbb Z\setminus\{0\}}\mathfrak g_{j\alpha},
$$
with absent <root spaces> understood to be zero. Its $H_\alpha$ weights are even, so every nontrivial <Irreducible Lie algebra representation> in it has even positive <highest weight> and one-dimensional weight-zero space. The action of $X_\alpha$ on its weight-zero space $\mathfrak h$ has image exactly $\mathbb C X_\alpha$, of dimension one. Therefore there is precisely one nontrivial irreducible summand, the already embedded adjoint $\mathfrak{sl}_2$ module of highest weight $2$. Thus $\dim\mathfrak g_\alpha=1$ and $2\alpha\notin R$. If $\beta=c\alpha$ is any root on the same real line, the integral numbers $\beta(H_\alpha)=2c$ and $\alpha(H_\beta)=2/c$ have product $4$. Hence $c$ is one of $\pm\tfrac12,\pm1,\pm2$; the half and double cases are excluded by applying the preceding argument to the appropriate root. This proves the <root-space reducedness lemma> and $R\cap\mathbb R\alpha=\{\pm\alpha\}$.
For a root $\beta$ not parallel to $\alpha$, the sum of <root spaces> $\bigoplus_{j\in\mathbb Z}\mathfrak g_{\beta+j\alpha}$ is stable under the root $\mathfrak{sl}_2$. Its integer $H_\alpha$ weights are symmetric under sign in each irreducible summand. Therefore the weight $-\beta(H_\alpha)$ also occurs, at the root $\beta-\beta(H_\alpha)\alpha$. The <Killing form> normalization gives
$$
\beta(H_\alpha)=\frac{2(\beta,\alpha)}{(\alpha,\alpha)},
$$
so this root is exactly $s_\alpha(\beta)$. For $\beta=\pm\alpha$ the reflection just swaps the two roots. This proves reflection invariance and the <Cartan integer> condition. All axioms of the reduced <crystallographic root system> have now been verified.
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