Morphism $f\colon X\to S$ is proper iff $f^{-1}(V_j)\to V_j$ is proper for some open cover $\{V_j\}$ of $S$? (Lemma 28.42.3 of Stacks Project) I was browsing the Stacks Project, and Lemma 28.42.3 says that a morphism $f\colon X\to S$ is a proper morphism if and only if there exists an open covering $S=\bigcup V_j$ such that $f^{-1}(V_j)\to V_j$ is proper for all $j$. However, the proof there is omitted.
Can anyone fill in this gap, or have a resource which explains it? Thanks.
 A: Let $\{V_j\to S\}$ be a Zariski cover of $S$. 
If $f:X\to S$ is proper, then the base-change $f_j=f|_{f^{-1}V_j} : f^{-1} V_j = X \times_S V_j \to V_j$ is proper.
Conversely, assume that $f_j: f^{-1}V_j \to V_j$ is proper for all $j$. Then $f_j$ is finite type and separated. In particular, since finite type and separated are local in the Zariski topology, we conclude that $f$ is finite type separated. (This is what Hoot is advising you to look up how to prove.) Therefore, it remains to show that $f$ is universally closed. If $T\to S$ is a morphism, then $\{T\times_S V_j\to T\}$ is a Zariski cover and the pull-back of $f$ to $T\times_S V_j$ is proper by base-change. Thus, to show that $f$ is universally closed, it suffices to show that $f$ is closed. (The same argument will then apply to $f:X\times_S T \to T$.
So let us show that $f:X\to S$ is closed. Let $Z$ be a closed subset of $X$. Let $f(Z)$ be the image of $Z$ in $S$. To conclude, it suffices to show that $f(Z)$ is closed.
Note that $f(Z)\cap V_j$ is closed for all $j$. Therefore, by a small lemma in topology, we conclude that $f(Z)$ is closed in $S$. QED
