[proofplan] The proof identifies both $X_H$ and $R_\alpha$ with the characteristic line field of the hypersurface $\Sigma$. Transversality of the Liouville vector field gives $dH(Y)\neq 0$, while regularity of the level set makes $X_H$ a nonzero tangent vector field spanning the kernel of $\omega|_{T\Sigma}$. The contact-type condition makes $\alpha=i^*(\iota_Y\omega)$ a contact form whose Reeb vector field spans the same line. The scalar factor is then fixed by the normalization $\alpha(R_\alpha)=1$, using the sign convention $\omega(X_H,\cdot)=dH$. [/proofplan]

[step:Use transversality to prove $dH(Y)$ is nowhere zero] Define the smooth function $\rho:\Sigma\to\mathbb R$ by \begin{align*} \rho(x)=dH_x(Y_x). \end{align*} Since $c$ is a regular value of $H$, the tangent space to the level hypersurface is \begin{align*} T_x\Sigma=\ker dH_x \end{align*} for every $x\in\Sigma$. The transversality hypothesis says $Y_x\notin T_x\Sigma$. Therefore $Y_x\notin\ker dH_x$, and hence \begin{align*} \rho(x)=dH_x(Y_x)\neq 0 \end{align*} for every $x\in\Sigma$. Thus $dH(Y)|_\Sigma$ is nowhere zero. [/step]

[step:Identify $X_H$ as the characteristic vector field of the regular energy hypersurface]For every $x\in\Sigma$, \begin{align*} dH_x((X_H)_x)=\omega_x((X_H)_x,(X_H)_x)=0, \end{align*} because $\omega_x$ is skew-symmetric. Hence $(X_H)_x\in T_x\Sigma$. Moreover, since $dH_x\neq 0$ and $\omega_x:T_xM\to T_x^*M$ is nondegenerate, $(X_H)_x\neq 0$. For every $v\in T_x\Sigma$, \begin{align*} \omega_x((X_H)_x,v)=dH_x(v)=0. \end{align*} Thus $(X_H)_x$ lies in \begin{align*} \ker(\omega_x|_{T_x\Sigma})=\{u\in T_x\Sigma:\omega_x(u,v)=0\text{ for every }v\in T_x\Sigma\}. \end{align*} By the characteristic-line statement for regular energy hypersurfaces, [citetheorem:10061], this kernel is the line spanned by $(X_H)_x$: \begin{align*} \ker(\omega_x|_{T_x\Sigma})=\mathbb R (X_H)_x. \end{align*}[/step]

[guided]The first point is that $X_H$ is actually tangent to the energy shell. The Hamiltonian vector field is defined by the convention \begin{align*} \omega(X_H,\cdot)=dH. \end{align*} Evaluating this one-form on $X_H$ itself gives \begin{align*} dH_x((X_H)_x)=\omega_x((X_H)_x,(X_H)_x). \end{align*} The right-hand side is zero because $\omega_x$ is skew-symmetric, so \begin{align*} dH_x((X_H)_x)=0. \end{align*} Since $\Sigma=H^{-1}(c)$ is a regular level set, its tangent space is \begin{align*} T_x\Sigma=\ker dH_x. \end{align*} Therefore $(X_H)_x\in T_x\Sigma$. Next, $X_H$ is nonzero on $\Sigma$. If $(X_H)_x=0$, then the one-form $\omega_x((X_H)_x,\cdot)$ would be zero. But this one-form is $dH_x$, and $dH_x\neq 0$ because $c$ is a regular value. This contradiction proves $(X_H)_x\neq 0$. Finally, if $v\in T_x\Sigma$, then $dH_x(v)=0$, and the Hamiltonian defining equation gives \begin{align*} \omega_x((X_H)_x,v)=dH_x(v)=0. \end{align*} Thus $(X_H)_x$ belongs to the characteristic kernel \begin{align*} \ker(\omega_x|_{T_x\Sigma})=\{u\in T_x\Sigma:\omega_x(u,v)=0\text{ for every }v\in T_x\Sigma\}. \end{align*} The theorem [citetheorem:10061] applies because $\Sigma=H^{-1}(c)$ is a regular energy hypersurface and $X_H$ is defined by $\omega(X_H,\cdot)=dH$. It identifies the characteristic kernel with the line generated by $(X_H)_x$: \begin{align*} \ker(\omega_x|_{T_x\Sigma})=\mathbb R (X_H)_x. \end{align*}[/guided]

[step:Identify the Reeb vector field with the same characteristic line] By the [contact-type restriction lemma](/theorems/10062), [citetheorem:10062], the form \begin{align*} \alpha=i^*(\iota_Y\omega) \end{align*} is a contact form on $\Sigma$, and its Reeb vector field satisfies \begin{align*} (R_\alpha)_x\in\ker(\omega_x|_{T_x\Sigma}) \end{align*} for every $x\in\Sigma$. Combining this with the previous step, for each $x\in\Sigma$ there is a unique scalar $a(x)\in\mathbb R$ such that \begin{align*} (R_\alpha)_x=a(x)(X_H)_x. \end{align*} [/step]

[step:Compute the scalar from the Reeb normalization] For every $x\in\Sigma$, \begin{align*} \alpha_x((X_H)_x)=(\iota_Y\omega)_x((X_H)_x)=\omega_x(Y_x,(X_H)_x). \end{align*} Using skew-symmetry of $\omega_x$ and the Hamiltonian convention, \begin{align*} \omega_x(Y_x,(X_H)_x)=-\omega_x((X_H)_x,Y_x)=-dH_x(Y_x). \end{align*} Thus \begin{align*} \alpha_x((X_H)_x)=-dH_x(Y_x). \end{align*} Since $(R_\alpha)_x=a(x)(X_H)_x$ and $\alpha((R_\alpha)_x)=1$, we get \begin{align*} 1=a(x)\alpha_x((X_H)_x)=-a(x)dH_x(Y_x). \end{align*} Because $dH_x(Y_x)\neq 0$, this gives \begin{align*} a(x)=-\frac{1}{dH_x(Y_x)}. \end{align*} Therefore, for every $x\in\Sigma$, \begin{align*} (R_\alpha)_x=-\frac{1}{dH_x(Y_x)}(X_H)_x. \end{align*} This is the desired reparametrisation formula. [/step]