The affine line is not proper

The multiplicative set $S$ contains no zero divisors of $k[t]$, and $k[t]$ is a domain by [F6], so [F5] shows that the localisation map $k[t]\to R$ is injective; hence $R$ is a domain and $(0)$ is a prime of $R$. Since $(t)\subseteq k[t]$ is a prime disjoint from $S$, [F4] shows that $R$ is a local ring with unique maximal ideal $\mathfrak{m}=tR$, and $t\in \mathfrak{m}$ is therefore not a unit of $R$. In particular $\mathrm{Spec}R$ contains the two points $(0)$ and $\mathfrak{m}$.

By [F9], ${\mathbf{A}}_k^1=\mathrm{Spec}k[x]$ over $S_0=\mathrm{Spec}k$, and $\mathrm{Spec}R\to S_0$ is the morphism induced by the field map $k\to R$. The canonical map $k[x]{\otimes }_{k}R\to R[x]$, $x^i\otimes r\mapsto r{x}^i$, is an isomorphism of $R$-algebras because $k[x]$ is the free $k$-module with basis $x^i$; hence [F7] identifies the base change ${\mathbf{A}}_k^1{\times }_{S_0}\mathrm{Spec}R$ with $\mathrm{Spec}R[x]$, with projection $q:\mathrm{Spec}R[x]\to \mathrm{Spec}R$ induced by $R\hookrightarrow R[x]$. By [F8] this $q$ is exactly the base change of ${\mathbf{A}}_k^1\to \mathrm{Spec}k$ along $\mathrm{Spec}R\to \mathrm{Spec}k$.

Let $Z=V(tx-1)\subseteq \mathrm{Spec}R[x]$ be the closed subset defined by the element $tx-1$. We compute its image. If $P\in Z$ is a prime containing $tx-1$ and if $t\in P\cap R$, then $tx-(tx-1)=1\in P$, a contradiction; so $t\mathrm{\notin }P\cap R$ and $P\cap R\in D(t)$. Conversely, let $\mathfrak{q}\in D(t)$ be a prime of $R$ and let $\psi :R[x]\to \mathrm{Frac}(R\mathrm{/}\mathfrak{q})$ be the $R$-algebra map sending $x$ to the inverse of the image of $t$, which is legitimate because $t\mathrm{\notin }\mathfrak{q}$. Its kernel $P$ is prime, contains $tx-1$, and satisfies $P\cap R=\mathfrak{q}$ because the composite $R\to \mathrm{Frac}(R\mathrm{/}\mathfrak{q})$ has kernel $\mathfrak{q}$. Hence the image of $Z$ under the base-changed projection $q$ of step 1.2 is exactly the basic open $D(t)\subseteq \mathrm{Spec}R$.

The subset $D(t)$ is open but not closed. It is nonempty because $(0)\in D(t)$ by step 1.1, and it is not all of $\mathrm{Spec}R$ because the maximal ideal $\mathfrak{m}=tR$ contains $t$ and so lies outside $D(t)$. If $D(t)$ were closed, then by [F3] it would equal $V(I)$ for the radical ideal $I$; from $(0)\in V(I)$ we get $I\subseteq (0)$, hence $I=0$, hence $V(I)=\mathrm{Spec}R\mathrm{\ne }D(t)$, a contradiction. Therefore $q$, whose image of the closed subset $Z$ is $D(t)$ by step 2.1, is not a closed map.

By [F2] and step 1.2, a nonclosed base-changed projection exhibits a failure of universal closedness of ${\mathbf{A}}_k^1\to \mathrm{Spec}k$, so that morphism is not universally closed and hence not proper by [F1]. The morphism is nevertheless of finite type and separated, being affine, so the failure is exactly in universal closedness. No choice principle is used: the prime witnessing nonclosedness is the maximal ideal $tR$, and the prime making $D(t)$ nonempty is $(0)$. Equivalently, the point $x=1\mathrm{/}t\in K$ has no $R$-lift, since a compatible $R$-point would give a $k$-algebra map $k[x]\to R$ with $x\mapsto 1\mathrm{/}t$ while $1\mathrm{/}t\mathrm{\notin }R$ because $t$ is not a unit of $R$ by step 1.1.