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Use the standard convention that a projective variety is nonempty and irreducible, over an algebraically closed field. Let $f:X\to {\mathbb{A}}^1$ be a regular function, and compose with the inclusion into ${\mathbb{P}}^1$. Its graph is closed in $X\times {\mathbb{P}}^1$: on an affine open of $X$, with homogeneous target coordinates $[T_0:T_1]$, it is given by $T_1-f(x)T_0=0$. This equation has no solution at $T_0=0$, and the local closedness statements glue.

Embed $X$ as a closed subvariety of ${\mathbb{P}}^m$. The graph is then closed in ${\mathbb{P}}^m\times {\mathbb{P}}^1$. Part 1, after swapping factors, makes its projected image $f(X)$ closed in ${\mathbb{P}}^1$. Since it avoids infinity, it is a proper closed subset of the projective line and hence finite. A continuous image of an irreducible space is irreducible, and a nonempty finite closed subset over an algebraically closed field is irreducible only if it is a singleton. HenceThis proves global regular functions on an irreducible projective variety are constant, using the closed-projection result rather than assuming it. If one allows disconnected projective algebraic sets instead of irreducible varieties, different components can carry different constants; irreducibility is the relevant convention here.