[proofplan] We fix a root $\alpha$ and use the Killing form to build the embedded $\mathfrak{sl}_2(k)$ generated by a nonzero vector in $\mathfrak{g}_{\alpha}$ and a suitably normalized vector in $\mathfrak{g}_{-\alpha}$. The root-string theorem supplies the two rank-one facts needed here: $\alpha$ is non-isotropic and $2\alpha$ is not a root. We then compare an arbitrary vector of $\mathfrak{g}_{\alpha}$ with the chosen root vector, subtract the component paired with $\mathfrak{g}_{-\alpha}$, and use the $\mathfrak{sl}_2$ commutator relation to force the remainder to vanish. [/proofplan]

[step:Use the Killing form to construct the $\mathfrak{sl}_2(k)$ attached to $\alpha$]Let \begin{align*} B: \mathfrak{g} \times \mathfrak{g} &\to k \\ (x,y) &\mapsto \operatorname{tr}(\operatorname{ad}_x \circ \operatorname{ad}_y) \end{align*} be the Killing form of $\mathfrak{g}$. Let $\mathfrak{h}\subset \mathfrak{g}$ be the Cartan subalgebra with root system $\Phi\subset \mathfrak{h}^*$. Since $\mathfrak{g}$ is semisimple and $\operatorname{char} k = 0$, [Cartan's criterion for semisimplicity](/theorems/???) gives that $B$ is nondegenerate. Moreover, the standard [Killing-form orthogonality of root spaces](/theorems/???) gives \begin{align*} B(\mathfrak{g}_{\beta},\mathfrak{g}_{\gamma}) = 0 \quad\text{unless}\quad \beta+\gamma=0, \end{align*} and the induced pairing \begin{align*} \mathfrak{g}_{\alpha} \times \mathfrak{g}_{-\alpha} &\to k \\ (u,v) &\mapsto B(u,v) \end{align*} is nondegenerate. The same standard result gives that the restriction \begin{align*} B|_{\mathfrak{h}\times\mathfrak{h}}: \mathfrak{h}\times\mathfrak{h} &\to k \end{align*} is nondegenerate. For a linear functional $\mu\in \mathfrak{h}^*$, define the weight space \begin{align*} \mathfrak{g}_{\mu}:=\{w\in \mathfrak{g}: [h,w]=\mu(h)w\text{ for every }h\in\mathfrak{h}\}. \end{align*} With this convention, $\mathfrak{g}_0=\mathfrak{h}$, and if $\mu\notin \Phi\cup\{0\}$ then $\mathfrak{g}_{\mu}=0$. Define $t_{\alpha} \in \mathfrak{h}$ to be the unique element satisfying \begin{align*} B(t_{\alpha},h)=\alpha(h) \quad\text{for every } h \in \mathfrak{h}. \end{align*} The [Root String Theorem](/theorems/???) for finite-dimensional semisimple Lie algebras over algebraically closed fields of characteristic $0$ implies $\alpha(t_{\alpha}) \ne 0$ and $2\alpha \notin \Phi$; here we use the root-string result only in a form that is independent of the one-dimensionality of root spaces. Choose $0 \ne x \in \mathfrak{g}_{\alpha}$. By nondegeneracy of the pairing between $\mathfrak{g}_{\alpha}$ and $\mathfrak{g}_{-\alpha}$, choose $y \in \mathfrak{g}_{-\alpha}$ such that \begin{align*} B(x,y)=\frac{2}{\alpha(t_{\alpha})}. \end{align*} For $u \in \mathfrak{g}_{\alpha}$ and $v \in \mathfrak{g}_{-\alpha}$, invariance of $B$ gives, for every $h \in \mathfrak{h}$, \begin{align*} B([u,v],h) = B(u,[v,h]) = B(u,\alpha(h)v) = \alpha(h)B(u,v) = B(B(u,v)t_{\alpha},h). \end{align*} Since $B|_{\mathfrak{h}\times\mathfrak{h}}$ is nondegenerate, this proves \begin{align*} [u,v]=B(u,v)t_{\alpha}. \end{align*} In particular, if \begin{align*} h_{\alpha}:=[x,y], \end{align*} then \begin{align*} h_{\alpha} = \frac{2}{\alpha(t_{\alpha})}t_{\alpha}. \end{align*} Therefore \begin{align*} [h_{\alpha},x] &= \alpha(h_{\alpha})x = 2x,\\ [h_{\alpha},y] &= -\alpha(h_{\alpha})y = -2y,\\ [x,y] &= h_{\alpha}. \end{align*} Thus the subalgebra \begin{align*} \mathfrak{s}:=\operatorname{span}_k\{x,h_{\alpha},y\} \end{align*} is isomorphic to $\mathfrak{sl}_2(k)$ under the correspondence $e \mapsto x$, $h \mapsto h_{\alpha}$, and $f \mapsto y$.[/step]

[guided]The role of the Killing form is to let us normalize a root vector and an opposite root vector so that they satisfy the exact $\mathfrak{sl}_2(k)$ relations. Define \begin{align*} B: \mathfrak{g} \times \mathfrak{g} &\to k \\ (x,y) &\mapsto \operatorname{tr}(\operatorname{ad}_x \circ \operatorname{ad}_y). \end{align*} Let $\mathfrak{h}\subset \mathfrak{g}$ be the Cartan subalgebra with root system $\Phi\subset \mathfrak{h}^*$. Because $\mathfrak{g}$ is semisimple and the field has characteristic $0$, [Cartan's criterion for semisimplicity](/theorems/???) says that $B$ is nondegenerate. The root space decomposition is orthogonal for $B$ in the following precise sense: \begin{align*} B(\mathfrak{g}_{\beta},\mathfrak{g}_{\gamma})=0 \quad\text{unless}\quad \beta+\gamma=0. \end{align*} Consequently, the [Killing-form orthogonality of root spaces](/theorems/???) implies that $B$ pairs $\mathfrak{g}_{\alpha}$ nondegenerately with $\mathfrak{g}_{-\alpha}$ and that \begin{align*} B|_{\mathfrak{h}\times\mathfrak{h}}: \mathfrak{h}\times\mathfrak{h} &\to k \end{align*} is nondegenerate. For a linear functional $\mu\in \mathfrak{h}^*$, define the weight space \begin{align*} \mathfrak{g}_{\mu}:=\{w\in \mathfrak{g}: [h,w]=\mu(h)w\text{ for every }h\in\mathfrak{h}\}. \end{align*} This definition makes sense for every $\mu$, not only for roots. In particular, $\mathfrak{g}_0=\mathfrak{h}$, and if $\mu\notin \Phi\cup\{0\}$ then $\mathfrak{g}_{\mu}=0$. We now encode the root $\alpha$ as an element of $\mathfrak{h}$. Define $t_{\alpha} \in \mathfrak{h}$ by \begin{align*} B(t_{\alpha},h)=\alpha(h) \quad\text{for every } h \in \mathfrak{h}. \end{align*} The [Root String Theorem](/theorems/???) gives two rank-one facts: $\alpha(t_{\alpha}) \ne 0$ and $2\alpha$ is not a root. We use this theorem only in a form proved independently of the one-dimensionality of root spaces. The first fact lets us normalize; the second will later imply that brackets of two vectors in $\mathfrak{g}_{\alpha}$ vanish. Choose a nonzero vector $x \in \mathfrak{g}_{\alpha}$. Since the pairing with $\mathfrak{g}_{-\alpha}$ is nondegenerate, we may choose $y \in \mathfrak{g}_{-\alpha}$ such that \begin{align*} B(x,y)=\frac{2}{\alpha(t_{\alpha})}. \end{align*} For arbitrary $u \in \mathfrak{g}_{\alpha}$ and $v \in \mathfrak{g}_{-\alpha}$, the bracket $[u,v]$ lies in $\mathfrak{g}_0=\mathfrak{h}$. To identify it, test against $h \in \mathfrak{h}$ using [invariance of the Killing form](/theorems/3808): \begin{align*} B([u,v],h) = B(u,[v,h]) = B(u,\alpha(h)v) = \alpha(h)B(u,v) = B(B(u,v)t_{\alpha},h). \end{align*} Nondegeneracy on $\mathfrak{h}$ gives \begin{align*} [u,v]=B(u,v)t_{\alpha}. \end{align*} Applying this to $u=x$ and $v=y$, we get \begin{align*} h_{\alpha}:=[x,y]=\frac{2}{\alpha(t_{\alpha})}t_{\alpha}. \end{align*} Then \begin{align*} [h_{\alpha},x] &= \alpha(h_{\alpha})x = 2x,\\ [h_{\alpha},y] &= -\alpha(h_{\alpha})y = -2y,\\ [x,y] &= h_{\alpha}. \end{align*} These are exactly the defining relations of $\mathfrak{sl}_2(k)$, so \begin{align*} \mathfrak{s}:=\operatorname{span}_k\{x,h_{\alpha},y\} \end{align*} is an embedded copy of $\mathfrak{sl}_2(k)$.[/guided]

[step:Subtract the component detected by the opposite root vector]Let $z \in \mathfrak{g}_{\alpha}$ be arbitrary. Define \begin{align*} \lambda:=\frac{B(z,y)}{B(x,y)} \qquad\text{and}\qquad z_0:=z-\lambda x \in \mathfrak{g}_{\alpha}. \end{align*} Then \begin{align*} B(z_0,y) = B(z,y)-\lambda B(x,y) = 0. \end{align*} Using the bracket formula from the previous step, \begin{align*} [z_0,y]=B(z_0,y)t_{\alpha}=0. \end{align*} Since $2\alpha \notin \Phi$ and $2\alpha\ne 0$, the definition of $\mathfrak{g}_{2\alpha}$ above gives $\mathfrak{g}_{2\alpha}=0$. The [root-space bracket rule](/theorems/???) therefore gives \begin{align*} [x,z_0]\in \mathfrak{g}_{2\alpha}=0. \end{align*} Finally, because $z_0 \in \mathfrak{g}_{\alpha}$ and $\alpha(h_{\alpha})=2$, \begin{align*} [h_{\alpha},z_0]=2z_0. \end{align*}[/step]

[guided]We now show that every vector in $\mathfrak{g}_{\alpha}$ is a scalar multiple of the fixed vector $x$. Let $z \in \mathfrak{g}_{\alpha}$. The vector $y \in \mathfrak{g}_{-\alpha}$ gives a linear functional on $\mathfrak{g}_{\alpha}$ by pairing with the Killing form. We remove from $z$ the same paired component that $x$ has by defining \begin{align*} \lambda:=\frac{B(z,y)}{B(x,y)} \qquad\text{and}\qquad z_0:=z-\lambda x. \end{align*} Here $B(x,y) \ne 0$ by the normalization chosen in the previous step. Since both $z$ and $x$ lie in $\mathfrak{g}_{\alpha}$, the vector $z_0$ also lies in $\mathfrak{g}_{\alpha}$. The definition of $\lambda$ gives \begin{align*} B(z_0,y) = B(z,y)-\lambda B(x,y) = 0. \end{align*} The bracket formula already proved says that for vectors in opposite root spaces, \begin{align*} [u,v]=B(u,v)t_{\alpha}. \end{align*} Applying this with $u=z_0$ and $v=y$ gives \begin{align*} [z_0,y]=0. \end{align*} We also need the bracket with $x$. Since $x,z_0 \in \mathfrak{g}_{\alpha}$, the [root-space bracket rule](/theorems/???) says that their bracket lies in $\mathfrak{g}_{2\alpha}$. The root-string theorem gives $2\alpha \notin \Phi$, and $2\alpha\ne 0$, so the definition of $\mathfrak{g}_{2\alpha}$ above gives $\mathfrak{g}_{2\alpha}=0$. Therefore \begin{align*} [x,z_0]=0. \end{align*} Finally, because $z_0$ lies in the $\alpha$-root space and $h_{\alpha}$ was normalized by $\alpha(h_{\alpha})=2$, we have \begin{align*} [h_{\alpha},z_0]=2z_0. \end{align*} Thus $z_0$ is killed by both opposite root operators $x$ and $y$, but has $h_{\alpha}$-weight $2$.[/guided]

[step:Use the $\mathfrak{sl}_2(k)$ commutator relation to force the remainder to vanish] Consider the adjoint representation \begin{align*} \rho: \mathfrak{s} &\to \mathfrak{gl}(\mathfrak{g}) \\ a &\mapsto \operatorname{ad}_a. \end{align*} The relations obtained above give \begin{align*} \rho(x)z_0=0, \qquad \rho(y)z_0=0, \qquad \rho(h_{\alpha})z_0=2z_0. \end{align*} Since $h_{\alpha}=[x,y]$, the representation identity gives \begin{align*} \rho(h_{\alpha}) = [\rho(x),\rho(y)] = \rho(x)\rho(y)-\rho(y)\rho(x). \end{align*} Applying this identity to $z_0$ yields \begin{align*} 2z_0 = \rho(h_{\alpha})z_0 = \rho(x)\rho(y)z_0-\rho(y)\rho(x)z_0 = 0. \end{align*} Because $\operatorname{char} k=0$, it follows that $z_0=0$. Hence $z=\lambda x$, so every element of $\mathfrak{g}_{\alpha}$ lies in $\operatorname{span}_k\{x\}$. Since $x \ne 0$, we conclude \begin{align*} \dim_k \mathfrak{g}_{\alpha}=1. \end{align*} This proves the theorem. [/step]