Assume first that $W \to Z$ is also separated. Let $U'$ be a scheme over $U$ and let $\sigma \in F(U')$. By Morphisms of Spaces, Lemma \ref{spaces-morphisms-lemma-section-immersion} the morphism $\sigma$ is a closed immersion. Moreover, $\sigma$ is \'etale by Properties of Spaces, Lemma \ref{spaces-properties-lemma-etale-permanence}. Hence $\sigma$ is also an open immersion, see Morphisms of Spaces, Lemma \ref{spaces-morphisms-lemma-etale-universally-injective-open}. In other words, $Z_\sigma = \sigma(Z_{U'}) \subset W_{U'}$ is an open subspace such that the morphism $Z_\sigma \to Z_{U'}$ is an isomorphism. In particular, the morphism $Z_\sigma \to U'$ is finite. Hence we obtain a transformation of functors $$ F \longrightarrow (W/U)_{fin}, \quad \sigma \longmapsto (U' \to U, Z_\sigma) $$ where $(W/U)_{fin}$ is the finite part of the morphism $W \to U$ introduced in More on Groupoids in Spaces, Section \ref{spaces-more-groupoids-section-finite}. It is clear that this transformation of functors is injective (since we can recover $\sigma$ from $Z_\sigma$ as the inverse of the isomorphism $Z_\sigma \to Z_{U'}$). By More on Groupoids in Spaces, Proposition \ref{spaces-more-groupoids-proposition-finite-algebraic-space} we know that $(W/U)_{fin}$ is an algebraic space \'etale over $U$. Hence to finish the proof in this case it suffices to show that $F \to (W/U)_{fin}$ is representable and an open immersion. To see this suppose that we are given a morphism of schemes $U' \to U$ and an open subspace $Z' \subset W_{U'}$ such that $Z' \to U'$ is finite. Then it suffices to show that there exists an open subscheme $U'' \subset U'$ such that a morphism $T \to U'$ factors through $U''$ if and only if $Z' \times_{U'} T$ maps isomorphically to $Z \times_{U'} T$. This follows from More on Morphisms of Spaces, Lemma \ref{spaces-more-morphisms-lemma-where-isomorphism} (here we use that $Z \to B$ is flat and locally of finite presentation as well as finite). Hence we have proved the lemma in case $W \to Z$ is separated as well as \'etale. \medskip\noindent In the general case we choose a separated scheme $W'$ and a surjective \'etale morphism $W' \to W$. Note that the morphisms $W' \to W$ and $W \to Z$ are separated as their source is separated. Denote $F'$ the functor associated to $W' \to Z \to U$ as in the lemma. In the first paragraph of the proof we showed that $F'$ is representable by an algebraic space \'etale over $U$. By Lemma \ref{lemma-surjection-space-of-sections} the map of functors $F' \to F$ is surjective for the \'etale topology on $\Sch/U$. Moreover, if $U'$ and $\sigma : Z_{U'} \to W_{U'}$ define a point $\xi \in F(U')$, then the fibre product $$ F'' = F' \times_{F, \xi} U' $$ is the functor on $\Sch/U'$ associated to the morphisms $$ W'_{U'} \times_{W_{U'}, \sigma} Z_{U'} \to Z_{U'} \to U'. $$ Since the first morphism is separated as a base change of a separated morphism, we see that $F''$ is an algebraic space \'etale over $U'$ by the result of the first paragraph. It follows that $F' \to F$ is a surjective \'etale transformation of functors, which is representable by algebraic spaces. Hence $F$ is an algebraic space by Bootstrap, Theorem \ref{bootstrap-theorem-final-bootstrap}. Since $F' \to F$ is an \'etale surjective morphism of algebraic spaces it follows that $F \to U$ is \'etale because $F' \to U$ is \'etale.