This is related to Ueno's Algebraic Geometry 3, Chpt 7, Example 7.33.
Let $k$ be a field $char(k)\neq 2$ and $f$ does not have multiple roots in algebraic closure. "$\pi: \operatorname{Spec}(k[x,y]/(y^2-f(x))=X\to \operatorname{Spec}(k)$ canonical projection map is not proper."
Q: Why is it obvious that $\pi$ is not proper map? Since $X$ is noetherian scheme and $\pi$ is of finite type, it suffices to check $\pi$ does not fulfill valuative criterion. The book will projectivize $X$ by homogenize defining equation to get proper morphism which is clear to me. However, it might happen that $\pi$ is already proper. To me, valuative criterion is more or less local lifting property by fixing a point upstairs to allow lifting a thickening of the point to upstairs.
Q': How do I see a general scheme morphism is not proper besides valuative criterion?
$\pi$ is affine, so if $\pi$ were proper, it would be finite (EGA2, 6.7.1 for instance, or you could prove it yourself). But $X$ is not finite over $\operatorname{Spec} k$.
For the general case, you may show that a morphism is not universally closed - for instance, this is the easiest way to show that $\Bbb A^n_R\to \operatorname{Spec} R$ isn't proper for any $n>0$.