 Proper closed-point fibers do not imply properness

The open immersionis 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 ${\mathbb{A}}_k^1$, 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 $\mathrm{Spec}k[[t]]$ over ${\mathbb{A}}_k^1$ 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.