= Proper closed-point fibers do not imply properness
The <open immersion>
$$
j:\mathbb A^1_k\setminus\{0\}\longrightarrow\mathbb A^1_k
$$
is of <finite type> and has a proper <scheme-theoretic fibre> over every <point>: a <point> over its <residue field> away from zero, and the empty <scheme> over zero. Nevertheless, $j$ is not a <proper morphism>. Its whole source is a closed <subset> of itself, while its <image> omits zero and is not closed in $\mathbb A^1_k$, so $j$ is not <universally closed>.
The failure is also visible in the <valuative criterion for properness>. Send the coordinate $x$ to $t$ in the <discrete valuation ring> $k[[t]]$. Its <fraction field> $k((t))$ gives a <point> of the punctured line because $t$ is invertible there. An extension to $\operatorname{Spec}k[[t]]$ over $\mathbb A^1_k$ would require $t$ to be invertible in $k[[t]]$, which it is not. Proper fibers alone do not control whether a family has a <limit> inside the source.
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