The morphism $f$ is representable and a universal homeomorphism, see Morphisms of Spaces, Section \ref{spaces-morphisms-section-universal-homeomorphisms}. \medskip\noindent We first prove that the functor is faithful. Suppose that $V', V$ are objects of $Y_{spaces, \etale}$ and that $a, b : V' \to V$ are distinct morphisms over $Y$. Since $V', V$ are \'etale over $Y$ the equalizer $$ E =  V' \times_{(a, b), V \times_Y V, \Delta_{V/Y}} V $$ of $a, b$ is \'etale over $Y$ also. Hence $E \to V'$ is an \'etale monomorphism (i.e., an open immersion) which is an isomorphism if and only if it is surjective. Since $X \to Y$ is a universal homeomorphism we see that this is the case if and only if $E_X = V'_X$, i.e., if and only if $a_X = b_X$. \medskip\noindent Next, we prove that the functor is fully faithful. Suppose that $V', V$ are objects of $Y_{spaces, \etale}$ and that $c : V'_X \to V_X$ is a morphism over $X$. We want to construct a morphism $a : V' \to V$ over $Y$ such that $a_X = c$. Let $a' : V'' \to V'$ be a surjective \'etale morphism such that $V''$ is a separated algebraic space. If we can construct a morphism $a'' : V'' \to V$ such that $a''_X = c \circ a'_X$, then the two compositions $$ V'' \times_{V'} V'' \xrightarrow{\text{pr}_i} V'' \xrightarrow{a''} V $$ will be equal by the faithfulness of the functor proved in the first paragraph. Hence $a''$ will factor through a unique morphism $a : V' \to V$ as $V'$ is (as a sheaf) the quotient of $V''$ by the equivalence relation $V'' \times_{V'} V''$. Hence we may assume that $V'$ is separated. In this case the graph $$ \Gamma_c \subset (V' \times_Y V)_X $$ is open and closed (details omitted). Since $X \to Y$ is a universal homeomorphism, there exists an open and closed subspace $\Gamma \subset V' \times_Y V$ such that $\Gamma_X = \Gamma_c$. The projection $\Gamma \to V'$ is an \'etale morphism whose base change to $X$ is an isomorphism. Hence $\Gamma \to V'$ is \'etale, universally injective, and surjective, so an isomorphism by Morphisms of Spaces, Lemma \ref{spaces-morphisms-lemma-etale-universally-injective-open}. Thus $\Gamma$ is the graph of a morphism $a : V' \to V$ as desired. \medskip\noindent Finally, we prove that the functor is essentially surjective. Suppose that $U$ is an object of $X_{spaces, \etale}$. We have to find an object $V$ of $Y_{spaces, \etale}$ such that $V_X \cong U$. Let $U' \to U$ be a surjective \'etale morphism such that $U' \cong V'_X$ and $U' \times_U U' \cong V''_X$ for some objects $V'', V'$ of $Y_{spaces, \etale}$. Then by fully faithfulness of the functor we obtain morphisms $s, t : V'' \to V'$ with $t_X = \text{pr}_0$ and $s_X = \text{pr}_1$ as morphisms $U' \times_U U' \to U'$. Using that $(\text{pr}_0, \text{pr}_1) : U' \times_U U' \to U' \times_S U'$ is an \'etale equivalence relation, and that $U' \to V'$ and $U' \times_U U' \to V''$ are universally injective and surjective we deduce that $(t, s) : V'' \to V' \times_S V'$ is an \'etale equivalence relation. Then the quotient $V = V'/V''$ (see Spaces, Theorem \ref{spaces-theorem-presentation}) is an algebraic space $V$ over $Y$. There is a morphism $V' \to V$ such that $V'' = V' \times_V V'$. Thus we obtain a morphism $V \to Y$ (see Descent on Spaces, Lemma \ref{spaces-descent-lemma-fpqc-universal-effective-epimorphisms}). On base change to $X$ we see that we have a morphism $U' \to V_X$ and a compatible isomorphism $U' \times_{V_X} U' = U' \times_U U'$, which implies that $V_X \cong U$ (by the lemma just cited once more). \medskip\noindent Pick a scheme $W$ and a surjective \'etale morphism $W \to Y$. Pick a scheme $U'$ and a surjective \'etale morphism $U' \to U \times_X W_X$. Note that $U'$ and $U' \times_U U'$ are schemes \'etale over $X$ whose structure morphism to $X$ factors through the scheme $W_X$. Hence by \'Etale Cohomology, Theorem \ref{etale-cohomology-theorem-topological-invariance} there exist schemes $V', V''$ \'etale over $W$ whose base change to $W_X$ is isomorphic to respectively $U'$ and $U' \times_U U'$. This finishes the proof.