Pick $f \in S_d$. Denote $s = \psi(f) \in \Gamma(Y, \mathcal{L})$. On the open set $Y_s$ where $s$ does not vanish multiplication by $s$ induces an isomorphism $\mathcal{O}_{Y_s} \to \mathcal{L}|_{Y_s}$, see Modules, Lemma \ref{modules-lemma-s-open}. We will denote the inverse of this map $x \mapsto x/s$, and similarly for powers of $\mathcal{L}$. Using this we define a ring map $\psi_{(f)} : S_{(f)} \to \Gamma(Y_s, \mathcal{O})$ by mapping the fraction $a/f^n$ to $\psi(a)/s^n$. By Schemes, Lemma \ref{schemes-lemma-morphism-into-affine} this corresponds to a morphism $\varphi_f : Y_s \to \Spec(S_{(f)}) = D_{+}(f)$. We also introduce the isomorphism $\alpha_f : \varphi_f^*\mathcal{O}_{D_{+}(f)}(d) \to \mathcal{L}|_{Y_s}$ which maps the pullback of the trivializing section $f$ over $D_{+}(f)$ to the trivializing section $s$ over $Y_s$. With this choice the commutativity of the diagram in the lemma holds with $Y$ replace by $Y_s$, $\varphi$ replaced by $\varphi_f$, and $\alpha$ replaced by $\alpha_f$; verification omitted. \medskip\noindent Suppose that $f' \in S_d$ is a second element, and denote $s' = \psi(f') \in \Gamma(Y, \mathcal{L})$. Then $Y_s \cap Y_{s'} = Y_{ss'}$ and similarly $D_{+}(f) \cap D_{+}(f') = D_{+}(ff')$. In Lemma \ref{lemma-ample-on-proj} we saw that $D_{+}(f') \cap D_{+}(f)$ is the same as the set of points of $D_{+}(f)$ where the section of $\mathcal{O}_X(d)$ defined by $f'$ does not vanish. Hence $\varphi_f^{-1}(D_{+}(f') \cap D_{+}(f)) = Y_s \cap Y_{s'} = \varphi_{f'}^{-1}(D_{+}(f') \cap D_{+}(f))$. On $D_{+}(f) \cap D_{+}(f')$ the fraction $f/f'$ is an invertible section of the structure sheaf with inverse $f'/f$. Note that $\psi_{(f')}(f/f') = \psi(f)/s' = s/s'$ and $\psi_{(f)}(f'/f) = \psi(f')/s = s'/s$. We claim there is a unique ring map $S_{(ff')} \to \Gamma(Y_{ss'}, \mathcal{O})$ making the following diagram commute $$ \xymatrix{ \Gamma(Y_s, \mathcal{O}) \ar[r] & \Gamma(Y_{ss'}, \mathcal{O}) & \Gamma(Y_{s, '} \mathcal{O}) \ar[l]\\ S_{(f)} \ar[r] \ar[u]^{\psi_{(f)}} & S_{(ff')} \ar[u] & S_{(f')} \ar[l] \ar[u]^{\psi_{(f')}} } $$ It exists because we may use the rule $x/(ff')^n \mapsto \psi(x)/(ss')^n$, which ``works'' by the formulas above. Uniqueness follows as $\text{Proj}(S)$ is separated, see Lemma \ref{lemma-proj-separated} and its proof. This shows that the morphisms $\varphi_f$ and $\varphi_{f'}$ agree over $Y_s \cap Y_{s'}$. The restrictions of $\alpha_f$ and $\alpha_{f'}$ agree over $Y_s \cap Y_{s'}$ because the regular functions $s/s'$ and $\psi_{(f')}(f)$ agree. This proves that the morphisms $\psi_f$ glue to a global morphism from $Y$ into $U_d \subset X$, and that the maps $\alpha_f$ glue to an isomorphism satisfying the conditions of the lemma. \medskip\noindent We still have to show the pair $(\varphi, \alpha)$ is unique. Suppose $(\varphi', \alpha')$ is a second such pair. Let $f \in S_d$. By the commutativity of the diagrams in the lemma we have that the inverse images of $D_{+}(f)$ under both $\varphi$ and $\varphi'$ are equal to $Y_{\psi(f)}$. Since the opens $D_{+}(f)$ are a basis for the topology on $X$, and since $X$ is a sober topological space (see Schemes, Lemma \ref{schemes-lemma-scheme-sober}) this means the maps $\varphi$ and $\varphi'$ are the same on underlying topological spaces. Let us use $s = \psi(f)$ to trivialize the invertible sheaf $\mathcal{L}$ over $Y_{\psi(f)}$. By the commutativity of the diagrams we have that $\alpha^{\otimes n}(\psi^d_{\varphi}(x)) = \psi(x) = (\alpha')^{\otimes n}(\psi^d_{\varphi'}(x))$ for all $x \in S_{nd}$. By construction of $\psi^d_{\varphi}$ and $\psi^d_{\varphi'}$ we have $\psi^d_{\varphi}(x) = \varphi^\sharp(x/f^n) \psi^d_{\varphi}(f^n)$ over $Y_{\psi(f)}$, and similarly for $\psi^d_{\varphi'}$. by the commutativity of the diagrams of the lemma we deduce that $\varphi^\sharp(x/f^n) = (\varphi')^\sharp(x/f^n)$. This proves that $\varphi$ and $\varphi'$ induce the same morphism from $Y_{\psi(f)}$ into the affine scheme $D_{+}(f) = \Spec(S_{(f)})$. Hence $\varphi$ and $\varphi'$ are the same as morphisms. Finally, it remains to show that the commutativity of the diagram of the lemma singles out, given $\varphi$, a unique $\alpha$. We omit the verification.