This proof is a repeat of the proof in the case of schemes, see Morphisms, Lemma \ref{morphisms-lemma-universally-closed-quasi-compact}. Let $f : X \to Y$ be a morphism of algebraic spaces over $S$. Assume that $f$ is not quasi-compact. Our goal is to show that $f$ is not universally closed. By Lemma \ref{lemma-quasi-compact-local} there exists an affine scheme $Z$ and a morphism $Z \to Y$ such that $Z \times_Y X \to Z$ is not quasi-compact. To achieve our goal it suffices to show that $Z \times_Y X \to Z$ is not universally closed, hence we may assume that $Y = \Spec(B)$ for some ring $B$. \medskip\noindent Write $X = \bigcup_{i \in I} X_i$ where the $X_i$ are quasi-compact open subspaces of $X$. For example, choose a surjective \'etale morphism $U \to X$ where $U$ is a scheme, choose an affine open covering $U = \bigcup U_i$ and let $X_i \subset X$ be the image of $U_i$. We will use later that the morphisms $X_i \to Y$ are quasi-compact, see Lemma \ref{lemma-quasi-compact-permanence}. Let $T = \Spec(B[a_i ; i \in I])$. Let $T_i = D(a_i) \subset T$. Let $Z \subset T \times_Y X$ be the reduced closed subspace whose underlying closed set of points is $|T \times_Y Z| \setminus \bigcup_{i \in I} |T_i \times_Y X_i|$, see Properties of Spaces, Lemma \ref{spaces-properties-lemma-reduced-closed-subspace}. (Note that $T_i \times_Y X_i$ is an open subspace of $T \times_Y X$ as $T_i \to T$ and $X_i \to X$ are open immersions, see Spaces, Lemmas \ref{spaces-lemma-base-change-immersions} and \ref{spaces-lemma-composition-immersions}.) Here is a diagram $$ \xymatrix{ Z \ar[r] \ar[rd] & T \times_Y X \ar[d]^{f_T} \ar[r]_q & X \ar[d]^f \\ & T \ar[r]^p & Y } $$ It suffices to prove that the image $f_T(|Z|)$ is not closed in $|T|$. \medskip\noindent We claim there exists a point $y \in Y$ such that there is no affine open neighborhood $V$ of $y$ in $Y$ such that $X_V$ is quasi-compact. If not then we can cover $Y$ with finitely many such $V$ and for each $V$ the morphism $Y_V \to V$ is quasi-compact by Lemma \ref{lemma-quasi-compact-permanence} and then Lemma \ref{lemma-quasi-compact-local} implies $f$ quasi-compact, a contradiction. Fix a $y \in Y$ as in the claim. \medskip\noindent Let $t \in T$ be the point lying over $y$ with $\kappa(t) = \kappa(y)$ such that $a_i = 1$ in $\kappa(t)$ for all $i$. Suppose $z \in |Z|$ with $f_T(z) = t$. Then $q(t) \in X_i$ for some $i$. Hence $f_T(z) \not \in T_i$ by construction of $Z$, which contradicts the fact that $t \in T_i$ by construction. Hence we see that $t \in |T| \setminus f_T(|Z|)$. \medskip\noindent Assume $f_T(|Z|)$ is closed in $|T|$. Then there exists an element $g \in B[a_i; i \in I]$ with $f_T(|Z|) \subset V(g)$ but $t \not \in V(g)$. Hence the image of $g$ in $\kappa(t)$ is nonzero. In particular some coefficient of $g$ has nonzero image in $\kappa(y)$. Hence this coefficient is invertible on some affine open neighborhood $V$ of $y$. Let $J$ be the finite set of $j \in I$ such that the variable $a_j$ appears in $g$. Since $X_V$ is not quasi-compact and each $X_{i, V}$ is quasi-compact, we may choose a point $x \in |X_V| \setminus \bigcup_{j \in J} |X_{j, V}|$. In other words, $x \in |X| \setminus \bigcup_{j \in J} |X_j|$ and $x$ lies above some $v \in V$. Since $g$ has a coefficient that is invertible on $V$, we can find a point $t' \in T$ lying above $v$ such that $t' \not \in V(g)$ and $t' \in V(a_i)$ for all $i \notin J$. This is true because $V(a_i; i \in I \setminus J) = \Spec(B[a_j; j\in J])$ and the set of points of this scheme lying over $v$ is bijective with $\Spec(\kappa(v)[a_j; j \in J])$ and $g$ restricts to a nonzero element of this polynomial ring by construction. In other words $t' \not \in T_i$ for each $i \not \in J$. By Properties of Spaces, Lemma \ref{spaces-properties-lemma-points-cartesian} we can find a point $z$ of $X \times_Y T$ mapping to $x \in X$ and to $t' \in T$. Since $x \not \in |X_j|$ for $j \in J$ and $t' \not \in T_i$ for $i \in I \setminus J$ we see that $z \in |Z|$. On the other hand $f_T(z) = t' \not \in V(g)$ which contradicts $f_T(Z) \subset V(g)$. Thus the assumption ``$f_T(|Z|)$ closed'' is wrong and we conclude indeed that $f_T$ is not closed as desired.