A faithful rational action with a fixed point forces affineness

First suppose $G$ connected. Replace the dense fixed open $V$ by $W=V\cap V^{-1}$. For any $g\in G(k)$, the dense opens $W$ and $g{W}^{-1}$ intersect; by [F3] choose $a$ in their intersection and write $g=ab$ with $a,b\in W$. Both $b\cdot x=x$ and $a\cdot (b\cdot x)=x$ are defined, so [F1] implies that the total action is defined at $(g,x)$ and has value $x$. Its regular domain is open. Its closed complement cannot meet $G\times \{x\}$, because any nonempty such intersection would contain a closed point by [F3]. Thus the domain contains all of $G\times \{x\}$. Since $G$ is smooth and hence reduced, its restriction to this subscheme equals the constant morphism $x$: on affine target charts their coordinate differences vanish on all closed points, hence vanish in the reduced coordinate ring.

Apply [F2] to obtain a closed immersion $G\hookrightarrow {\mathrm{GL}}_N$ for some finite $N$. General linear groups are affine, and a closed subscheme of an affine scheme is affine, so $G$ is affine. For arbitrary smooth $G$, its identity component is open and closed: smooth local rings are domains, so irreducible components are disjoint, and translations identify the connected components. The rational action restricted to the identity component is scheme-faithful and its dense fixed open is nonempty, so the preceding argument makes that component affine. There are finitely many components and each is a translate of it by a rational point, available by [F3]. Their disjoint union is affine, the spectrum of the finite product of their coordinate rings. Hence $G$ is affine. AC is inherited from [F2] and the geometric suppliers.