Note that the assertion makes sense by the result of Lemma \ref{lemma-finite-diagonal} on the diagonal of $(X/Y)_{fin}$, see Spaces, Lemma \ref{spaces-lemma-representable-diagonal}. Let $V$ be a scheme and let $V \to Y$ be a surjective \'etale morphism. By Lemma \ref{lemma-finite-pullback} the morphism $(V \times_Y X/V)_{fin} \to (X/Y)_{fin}$ is a base change of the map $V \to Y$ and hence is surjective and \'etale, see Spaces, Lemma \ref{spaces-lemma-base-change-representable-transformations-property}. Hence it suffices to prove the lemma for $(V \times_Y X/V)_{fin}$. (Here we implicitly use that the composition of representable, surjective, and \'etale transformations of functors is again representable, surjective, and \'etale, see Spaces, Lemmas \ref{spaces-lemma-composition-representable-transformations} and \ref{spaces-lemma-composition-representable-transformations-property}, and Morphisms, Lemmas \ref{morphisms-lemma-composition-surjective} and \ref{morphisms-lemma-composition-etale}.) Note that the properties of being separated and locally quasi-finite are preserved under base change, see Morphisms of Spaces, Lemmas \ref{spaces-morphisms-lemma-base-change-separated} and \ref{spaces-morphisms-lemma-base-change-quasi-finite}. Hence $V \times_Y X \to V$ is separated and locally quasi-finite as well, and by Morphisms of Spaces, Proposition \ref{spaces-morphisms-proposition-locally-quasi-finite-separated-over-scheme} we see that $V \times_Y X$ is a scheme as well. Thus we may assume that $f : X \to Y$ is a separated and locally quasi-finite morphism of schemes. \medskip\noindent Pick a point $y \in Y$. Pick $x_1, \ldots, x_n \in X$ points lying over $y$. Pick an \'etale neighbourhood $a : (U, u) \to (Y, y)$ and a decomposition $$ U \times_S X = W \amalg \ \coprod\nolimits_{i = 1, \ldots, n} \ \coprod\nolimits_{j = 1, \ldots, m_j} V_{i, j} $$ as in More on Morphisms, Lemma \ref{more-morphisms-lemma-etale-splits-off-quasi-finite-part-technical-variant}. Pick any subset $$ I \subset \{(i, j) \mid 1 \leq i \leq n, \ 1 \leq j \leq m_i\}. $$ Given these choices we obtain a pair $(a, Z)$ with $Z = \bigcup_{(i, j) \in I} V_{i, j}$ which satisfies conditions \ref{equation-finite-conditions}. In other words we obtain a morphism $U \to (X/Y)_{fin}$. The construction of this morphism depends on all the things we picked above, so we should really write $$ U(y, n, x_1, \ldots, x_n, a, I) \longrightarrow (X/Y)_{fin} $$ This morphism is \'etale by Lemma \ref{lemma-finite-criterion-etale}. \medskip\noindent Claim: The disjoint union of all of these is surjective onto $(X/Y)_{fin}$. It is clear that if the claim holds, then the lemma is true. \medskip\noindent To show surjectivity we have to show the following (see Spaces, Remark \ref{spaces-remark-warning}): Given a scheme $T$ over $S$, a point $t \in T$, and a map $T \to (X/Y)_{fin}$ we can find a datum $(y, n, x_1, \ldots, x_n, a, I)$ as above such that $t$ is in the image of the projection map $$ U(y, n, x_1, \ldots, x_n, a, I) \times_{(X/Y)_{fin}} T \longrightarrow T. $$ To prove this we may clearly replace $T$ by $\Spec(\overline{\kappa(t)})$ and $T \to (X/Y)_{fin}$ by the composition $\Spec(\overline{\kappa(t)}) \to T \to (X/Y)_{fin}$. In other words, we may assume that $T$ is the spectrum of an algebraically closed field. \medskip\noindent Let $T = \Spec(k)$ be the spectrum of an algebraically closed field $k$. The morphism $T \to (X/Y)_{fin}$ is given by a pair $(T \to Y, Z)$ satisfying conditions \ref{equation-finite-conditions}. Here is a picture: $$ \xymatrix{ & Z \ar[d] \ar[r] & X \ar[d] \\ \Spec(k) \ar@{=}[r] & T \ar[r] & Y } $$ Let $y \in Y$ be the image point of $T \to Y$. Since $Z$ is finite over $k$ it has finitely many points. Thus there exist finitely many points $x_1, \ldots, x_n \in X$ such that the image of $Z$ in $X$ is contained in $\{x_1, \ldots, x_n\}$. Choose $a : (U, u) \to (Y, y)$ adapted to $y$ and $x_1, \ldots, x_n$ as above, which gives the diagram $$ \xymatrix{ W \amalg \ \coprod\nolimits_{i = 1, \ldots, n} \ \coprod\nolimits_{j = 1, \ldots, m_j} V_{i, j} \ar[d] \ar[r] & X \ar[d] \\ U \ar[r] & Y. } $$ Since $k$ is algebraically closed and $\kappa(y) \subset \kappa(u)$ is finite separable we may factor the morphism $T = \Spec(k) \to Y$ through the morphism $u = \Spec(\kappa(u)) \to \Spec(\kappa(y)) = y \subset Y$. With this choice we obtain the commutative diagram: $$ \xymatrix{ Z \ar[d] \ar[r] & W \amalg \ \coprod\nolimits_{i = 1, \ldots, n} \ \coprod\nolimits_{j = 1, \ldots, m_j} V_{i, j} \ar[d] \ar[r] & X \ar[d] \\ \Spec(k) \ar[r] & U \ar[r] & Y } $$ We know that the image of the left upper arrow ends up in $\coprod V_{i, j}$. Recall also that $Z$ is an open subscheme of $\Spec(k) \times_Y X$ by definition of $(X/Y)_{fin}$ and that the right hand square is a fibre product square. Thus we see that $$ Z \subset \coprod\nolimits_{i = 1, \ldots, n}\ \coprod\nolimits_{j = 1, \ldots, m_j} \Spec(k) \times_U V_{i, j} $$ is an open subscheme. By construction (see More on Morphisms, Lemma \ref{more-morphisms-lemma-etale-splits-off-quasi-finite-part-technical-variant}) each $V_{i, j}$ has a unique point $v_{i, j}$ lying over $u$ with purely inseparable residue field extension $\kappa(u) \subset \kappa(v_{i, j})$. Hence each scheme $\Spec(k) \times_U V_{i, j}$ has exactly one point. Thus we see that $$ Z = \coprod\nolimits_{(i, j) \in I} \Spec(k) \times_U V_{i, j} $$ for a unique subset $I \subset \{(i, j) \mid 1 \leq i \leq n, \ 1 \leq j \leq m_i\}$. Unwinding the definitions this shows that $$ U(y, n, x_1, \ldots, x_n, a, I) \times_{(X/Y)_{fin}} T $$ with $I$ as found above is nonempty as desired.