A nonclosed open immersion is not proper

The ring $R=k[t]$ is an integral domain by [F5], so $(0)$ is a prime ideal and is a point of $X=\mathrm{Spec}R={\mathbf{A}}_k^1$. The indeterminate satisfies $t\mathrm{\ne }0$ in $k[t]$, so $t\mathrm{\notin }(0)$ and therefore $(0)\in D(t)$.

The principal ideal $(t)$ equals $\{f\in k[t]:f(0)=0\}$. For $f=tg$ one has $f(0)=0\cdot g(0)=0$, so $(t)$ is contained in that set and in particular $t\in (t)$ by [F4]; conversely, if $f(0)=0$ then the constant coefficient of $f$ vanishes, so writing $f$ as its finite coefficient sum gives $f=t\cdot ({\sum }_{i\ge 1}{a}_i{t}^{i-1})\in (t)$. Hence $(t)$ is a prime ideal: it is proper because $1\mathrm{\notin }(t)$, as $\mathrm{ev}(1)=1\mathrm{\ne }0$; and if $fg\in (t)$ then $f(0)g(0)=\mathrm{ev}(f)\mathrm{ev}(g)=\mathrm{ev}(fg)=0$, so $f(0)=0$ or $g(0)=0$ because $k$ is an integral domain, and then $f\in (t)$ or $g\in (t)$. Thus $(t)$ is a point of $X$ and $t\in (t)$, so $(t)\mathrm{\notin }D(t)$.

The set $D(t)=X\setminus V((t))$ is open in $X$ by [F2], and the inclusion $j:D(t)\to X$ of this open subscheme is an open immersion by [F7] whose image is exactly the subset $D(t)\subseteq X$. By steps 1.1 and 1.2 that image contains $(0)$ and omits $(t)$, so it is a nonempty proper subset of $X$.

The image $D(t)$ is not closed in $X$. If it were closed, then, the closed subsets of $X$ being the vanishing sets [F2], there would be an ideal $I\subseteq k[t]$ with $D(t)=V(I)$; by [F5] the ideal is principal, $I=(g)$, so $D(t)=V(g)$. Since $(0)\in D(t)$ we get $g\in (0)$ by [F3], that is $g=0$, and then $D(t)=V((0))=X$ by [F3], contradicting the properness of the image recorded in step 2.1.

It follows that $j$ is not a closed map of topological spaces: the whole source $D(t)$ is a closed subset of the source, and its image under $j$ is $D(t)$, which is not closed in $X$ by step 3.1.

Suppose $j$ were universally closed. Taking the identity morphism $X\to X$ as test morphism, the definition [F8] would make the base-changed projection $D(t){\times }_{X}X\to X$ a closed map. By [F9] the scheme $D(t){\times }_{X}X$ is projection-compatibly isomorphic to $D(t)$, and under this isomorphism the projection corresponds to $j$, so $j$ would be a closed map, contradicting step 4.1. Therefore $j$ is not universally closed.

Since properness requires universal closedness by [F8], the open immersion $j$ is not proper; directly, if $j$ were proper then [F8] would make its image closed, contradicting step 3.1. So the claim that every open immersion is proper is refuted. The argument applies to every field, including finite fields and fields of characteristic two, since the two points $(0)$ and $(t)$ of $X$ exist for every field; both are exhibited explicitly and no choice principle is used, the only inputs being the definitional closedness description of the Zariski topology, the principal-ideal property of $k[t]$, and the elementary computation $(t)=\{f:f(0)=0\}$ of step 1.2.