Finite morphisms are closed with finite fibres

First the affine case: let $Y$ have coordinate ring $A=k[Y]$, let $X$ have coordinate ring $B=k[X]$, let $A\to B$ be the pullback of $f$, and assume $B$ is a finite $A$-module; let $Z\subseteq X$ be closed with radical ideal $J=I(Z)\subseteq B$. Then $f(Z)=V(J\cap A)$, which is closed. Indeed, if $z\in Z$ and $g\in J\cap A$, then $f^{\ast }(g)=g\circ f\in J$ vanishes at $z$, so $g(f(z))=0$, giving $f(Z)\subseteq V(J\cap A)$. Conversely let $y\in V(J\cap A)$ and let $\mathfrak{m}={\mathfrak{m}}_y\subseteq A$ be the corresponding maximal ideal, which contains $J\cap A$. The induced map $A\mathrm{/}(J\cap A)\to B\mathrm{/}J$ is integral, because a monic integrality equation for $b\in B$ over $A$ reduces modulo $J\cap A$ to a monic equation for the class of $b$; and $\bar{\mathfrak{m}}=\mathfrak{m}\mathrm{/}(J\cap A)$ is maximal. By lying over [F3] there is a prime $\bar{\mathfrak{q}}\subseteq B\mathrm{/}J$ contracting to $\bar{\mathfrak{m}}$, and $\bar{\mathfrak{q}}$ is maximal by [F4]; it is the image of a maximal ideal $\mathfrak{q}\subseteq B$ containing $J$. By [F5] the maximal ideal $\mathfrak{q}$ is ${\mathfrak{m}}_z$ for a point $z\in X$, and $J\subseteq \mathfrak{q}$ with $Z=V(J)$ gives $z\in Z$. Finally $\mathfrak{q}\cap A=\mathfrak{m}$, since both ideals contain $J\cap A$ and have the same image in $A\mathrm{/}(J\cap A)$; as contraction along the pullback is precomposition, $\mathfrak{q}\cap A=\{g\in A:g\circ f\in {\mathfrak{m}}_z\}=\{g\in A:g(f(z))=0\}={\mathfrak{m}}_{f(z)}$. Hence ${\mathfrak{m}}_{f(z)}={\mathfrak{m}}_y$, so $f(z)=y$ by the bijectivity in [F5], and $y\in f(Z)$.

Finite fibres: let $y\in Y$ and choose an index $i$ with $y\in U_i$; then $f^{-1}(y)\subseteq f^{-1}(U_i)$, and the restriction $f_i\text{\hspace{0.17em}⁣}:f^{-1}(U_i)\to U_i$ is a module-finite morphism of affine classical algebraic sets, because $B_i$ is a finite $A_i$-module by the choice of the cover [F1, F2]. By [F6] the map $f_i$ is quasi-finite, so its fibre $f_i^{-1}(y)=f^{-1}(y)$ is a finite set.

Now the general closedness statement. Let $Z\subseteq X$ be closed. A subset of $Y$ is closed exactly when its traces in the members of the open cover $U_1,\dots ,U_n$ are closed, and $f(Z)\cap U_i=f_i(Z\cap f^{-1}(U_i))$, where $Z\cap f^{-1}(U_i)$ is closed in the affine variety $f^{-1}(U_i)$. By step 1.1 applied to the affine map $f_i$, each $f(Z)\cap U_i$ is closed in $U_i$. Hence $f(Z)$ is closed in $Y$, so $f$ is closed.

Combining the two halves: every closed subset of $X$ has closed image by step 2.1, and every fibre of $f$ is finite by step 1.2. No characteristic, flatness, separability, or normality hypothesis entered either argument, and the only choice principle used is the Axiom of Choice through [F3] and [F5].