I am trying to show the following: if $X$ is an integral proper $k$-scheme, $k$ a field, then $O_X(X)$ is a finite field extension of $k$.
I have succeeded to show that $O_X(X)$ is a field but I don't see why it must be a finite field extension.
(To show it is a field I used that a global section s corresponds to a morphism $X \to \operatorname{Spec} k[x]$, one can show the image is a closed point, so if $s \neq 0$ there is an irreducible polynomial $g \in k[x]$ such that $g(s)=0$, so it is invertible.)
I would like to avoid using cohomology/Grothendieck's finiteness result for proper morphisms. a similar question was asked here but I am not assuming $X$ is geometrically integral.
The solution has been worked out in the comments. As restriction maps for the structure sheaf on an integral scheme are injective, $O_X(X)$ embeds in to $O_X(\operatorname{Spec} A)=A$ for $\operatorname{Spec} A$ any affine open subscheme of $X$. Let $\mathfrak{m}$ be a maximal ideal of $A$. Then $O_X(X)\subset A$ does not intersect $\mathfrak{m}$, so the map from $O_X(X)$ to its image in $A/\mathfrak{m}$ is an injection. Thus $O_X(X)$ embeds in to a residue field of a finite type scheme over $k$, and all such residue fields are finite extensions of $k$ by Zariski's lemma.