Every complex affine algebraic action has a finite-dimensional equivariant closed embedding

Choose finitely many algebra generators of $A$ by F1 and put them in a finite-dimensional rational submodule $W$ by F2. Then $W$ generates $A$. Choose a finite basis $w_1,\dots ,w_N$ of $W$. Its action has regular matrix entries; the dual action has transpose-inverse matrix, whose entries are regular because inversion $g\mapsto g^{-1}$ is a morphism. Thus $W^{\ast }$ is a finite-dimensional rational module.

The coordinate map $\pi :\mathbb{C}[z_1,\dots ,z_N]\to A$, $z_i\mapsto w_i$, is surjective. Its kernel $I$ is radical since $A$ is reduced. F4 identifies $\mathbb{C}[V(I)]$ with $\mathbb{C}[z_1,\dots ,z_N]\mathrm{/}I$, and $\pi$ identifies that quotient with $A$. F3 gives mutually inverse morphisms $X\leftrightarrow V(I)$ from this algebra isomorphism. The morphism to $W^{\ast }\cong {\mathbb{A}}^N$ is precisely evaluation because its coordinates are $w_i(x)$; hence evaluation is an isomorphism onto the closed set $V(I)$. If $X$ is empty, $A=0$, and $I$ is the unit ideal so the conclusion still holds.

For $g\in G$, $x\in X$ and $w\in W$, $(g\mathrm{e}\mathrm{v}(x))(w)=\mathrm{e}\mathrm{v}(x)(g^{-1}w)=(g^{-1}w)(x)=w(gx)=\mathrm{e}\mathrm{v}(gx)(w)$ by the inverse-pullback action on functions. This proves equivariance and invariance of the image. The only choice beyond finite-dimensional selection is the AC inherited by F3 and F4 in step 2.1; local finiteness itself needs no infinite basis.