If $X$ is an integral projective $k$-scheme and $X(k) \neq \emptyset$, I would like to prove that the global sections $H^0(X,\mathcal{O}_X)=k$.
My approach so far has been the following: I know that $X(k)$ being non empty implies that there exists a point $x \in X$ such that standard inclusion $k \rightarrow k(x)$ is a bijection ($k(x)$ being the residue field at x). Let $\operatorname{Spec}(A) \subset X$ be an affine open which contains $x$. Then the restriction from $\mathcal{O}_X(X) \rightarrow A$ is injective (restrictions of the structure sheaf are injective on integral schemes or using that the global sections is a field). It is my understanding that it would be enough then to prove that the composition map $\mathcal{O}_X(X) \rightarrow A \rightarrow k(x)$ is an isomorphism but I am not sure how to proceed with that argument.
You're almost there.
First, $\mathcal{O}_X(X)$ is an algebraic field extension of $k$: global sections correspond to $k$-morphisms $X\to\Bbb A^1_k$, and the image of $X$ must be a point (image of proper scheme is proper, image of irreducible space is irreducible, only proper irreducible subsets of $\Bbb A^1_k$ are closed points). Therefore we have a nonzero irreducible polynomial $f(t)\in k[t]$ which vanishes on our global section.
Next, $\mathcal{O}_X(X)$ embeds in $A$, as you've pointed out. The composite map $\mathcal{O}_X(X)\to k(x)$ is then a map of fields, so it must be injective. On the other hand, we have an injective morphism of $k$-algebras with target $k$, and this must be an isomorphism.