It is clear that (1) implies (2). Suppose that $|X \times_Y Z| \to |Z|$ is not closed for some morphism of algebraic spaces $Z \to Y$ over $S$. This means that there exists some closed subset $T \subset |X \times_Y Z|$ such that $\Im(T \to |Z|)$ is not closed. Pick $z \in |Z|$ in the closure of the image of $T$ but not in the image. Apply Lemma \ref{lemma-separate}. We find an \'etale neighbourhood $(V, v) \to (Z, z)$, a commutative diagram $$ \xymatrix{ V \ar[d] \ar[r]_a & Z' \ar[d]^b \\ Z \ar[r]^g & Y, } $$ and a closed subset $T' \subset |X \times_Y Z'|$ such that \begin{enumerate} \item the morphism $b : Z' \to Y$ is locally of finite presentation, \item with $z' = a(v)$ we have $z' \not \in \Im(T' \to |Z'|)$, and \item the inverse image of $T$ in $|X \times_Y V|$ maps into $T'$ via $|X \times_Y V| \to |X \times_Y Z'|$. \end{enumerate} We claim that $z'$ is in the closure of $\Im(T' \to |Z'|)$ which implies that $|X \times_Y Z'| \to |Z'|$ is not closed. The claim shows that (2) implies (1). To see the claim is true we suggest the reader contemplate the following commutative diagram $$ \xymatrix{ X \times_Y Z \ar[d] & X \times_Y V \ar[l] \ar[d] \ar[r] & X \times_Y Z' \ar[d] \\ Z & V \ar[l] \ar[r]^a & Z' } $$ Let $T_V \subset |X \times_Y V|$ be the inverse image of $T$. By Properties of Spaces, Lemma \ref{spaces-properties-lemma-points-cartesian} the image of $T_V$ in $|V|$ is the inverse image of the image of $T$ in $|Z|$. Then since $z$ is in the closure of the image of $T \to |Z|$ and since $|V| \to |Z|$ is open, we see that $v$ is in the closure of the image of $T_V \to |V|$. Since the image of $T_V$ in $|X \times_Y Z'|$ is contained in $|T'|$ it follows immediately that $z' = a(v)$ is in the closure of the image of $T'$. \medskip\noindent It is clear that (1) implies (3). Let $V \to Y$ be as in (3). If we can show that $X \times_Y V \to V$ is universally closed, then $f$ is universally closed by Morphisms of Spaces, Lemma \ref{spaces-morphisms-lemma-universally-closed-local}. Thus it suffices to show that $f : X \to Y$ satisfies (2) if $f$ is a quasi-compact morphism of algebraic spaces, $Y$ is a scheme, and $|\mathbf{A}^n \times X| \to |\mathbf{A}^n \times Y|$ is closed for all $n$. Let $Z \to Y$ be locally of finite presentation. We have to show the map $|X \times_Y Z| \to |Z|$ is closed. This question is \'etale local on $Z$ hence we may assume $Z$ is affine (some details omitted). Since $Y$ is a scheme, $Z$ is affine, and $Z \to Y$ is locally of finite presentation we can find an immersion $Z \to \mathbf{A}^n \times Y$, see Morphisms, Lemma \ref{morphisms-lemma-quasi-affine-finite-type-over-S}. Consider the cartesian diagram $$ \vcenter{ \xymatrix{ X \times_Y Z \ar[d] \ar[r] & \mathbf{A}^n \times X \ar[d] \\ Z \ar[r] & \mathbf{A}^n \times Y } } \quad \begin{matrix} \text{inducing the} \\ \text{cartesian square} \end{matrix} \quad \vcenter{ \xymatrix{ |X \times_Y Z| \ar[d] \ar[r] & |\mathbf{A}^n \times X| \ar[d] \\ |Z| \ar[r] & |\mathbf{A}^n \times Y| } } $$ of topological spaces whose horizontal arrows are homeomorphisms onto locally closed subsets (Properties of Spaces, Lemma \ref{spaces-properties-lemma-subspace-induced-topology}). Thus every closed subset $T$ of $|X \times_Y Z|$ is the pullback of a closed subset $T'$ of $|\mathbf{A}^n \times Y|$. Since the assumption is that the image of $T'$ in $|\mathbf{A}^n \times X|$ is closed we conclude that the image of $T$ in $|Z|$ is closed as desired.