Let $r\in\mathbb{Z}_{> 0}$, let $k$ be a field and $X=\mathbb{P}^{r}_{k}$ and $S=k[X_{0},...,X_{r}]$.
By definition we have that the function field is defined as $K(X)=\mathcal{O}_{X,\eta}$ where $\eta$ is the unique generic point of $X$. I want to show that this function field can be identified with the degree zero elements of the fraction field of $S$.
Notice that we can take an arbitrary affine open $U_{i}=\operatorname{Spec}(R_{i})$ where $R_{i}=k[...,X_{ji},...]_{i=0,...,r, j\neq i}$. Then the unique generic point corresponds to $(0)\in \operatorname{Spec}(R_{i})$ and we have $K(X)=\mathcal{O}_{U_{i},(0)}$. Note that the structure sheaf on $U_{i}$ is given by $\widetilde{(S_{X_{i}})_{0}}$ and thus we find that $K(X)\cong ((S_{X_{i}})_{0})_{(0)} = \operatorname{Frac}((S_{X_{i}})_{0})$.
From here I don't know how to finish the proof.
Any help would be appreciated!
Maybe you don't have to work in the affine piece $D_+(X_i)$.
Denote by $A = k[X_0,\dots,X_r]$. The generic point of $\mathrm{Proj} A$ is exact the zero ideal $0$ of $A$.
One way to see this: The Zariski topology of $\mathrm{Proj} A$ coincides with the induced topology of the Zariski topology of $\mathrm{Spec} A$. Hence, $\mathrm{Cls}_{\mathrm{Proj}A} 0 = \mathrm{Cls}_{\mathrm{Spec}A} 0 \cap \mathrm{Proj}A = V(0) \cap \mathrm{Proj}A = \mathrm{Spec}A \cap \mathrm{Proj}A = \mathrm{Proj}A$ where $\mathrm{Cls}_{\mathrm{Proj}A}0$ means taking closure of $0$ in $\mathrm{Proj}A$.
Let $\mathfrak{p} = 0$. So, $K(\mathrm{ProjA}) = A_{(\mathfrak{p})}$ degree zero pieces of the localization $A_\mathfrak{p} = k(X_0,\dots,X_r)$.