There is an elementary proof of the result "universally closed + affine $\Rightarrow$ integral" that I learnt from Olivier's paper "Going up along absolutely flat morphisms." In fact, it's so simple, I can present it here.

**Observation 1**: Say $\phi:A \to B$ is an injective ring map that is closed on $\mathrm{Spec}$. Then $\phi^{-1}(B^\ast) = A^\ast$.

*(This proof was edited and corrected to reflect xuhan's comment.)*

*Proof*: Fix $a \in A$ with $\phi(a) \in B^\ast$. We must show $a \in A^\ast$ or, equivalently, $a$ is non-zero in the residue field $\kappa(\mathfrak{p})$ of $A$ at any prime $\mathfrak{p} \in \mathrm{Spec}(A)$. Note that the last statement is clearly true if $\mathfrak{p}$ lies in the image $Z$ of $\mathrm{Spec}(\phi)$. So it suffices to show $Z = \mathrm{Spec}(A)$. By closedness, $Z = V(I)$ for some ideal $I \subset A$ (set-theoretically). Localizing at any prime $\mathfrak{p}$ shows $I \subset \mathfrak{p}$ by the injectivity hypothesis. Then $I$ is contained in all primes of $A$, so it contains only nilpotents, and hence $Z = \mathrm{Spec}(A)$.


**Observation 2**: Say $\phi:A \to B$ is an injective ring map, and $\phi[T]:A[T] \to B[T]$ is closed on $\mathrm{Spec}$. Then $\phi$ is integral.

*Proof*: Fix some $f \in B$, and consider the surjective map $B[T] \to B[\frac{1}{f}]$ given by $T \mapsto \frac{1}{f}$. If we write $C \subset B[\frac{1}{f}]$ for the image of the composite $A[T] \to B[T] \to B[\frac{1}{f}]$, then $C \to B[\frac{1}{f}]$ is an injective ring map that is closed on $\mathrm{Spec}$. The image of $T$ in $C$ becomes a unit in $B[\frac{1}{f}]$, and hence must be a unit on $C$ by Observation 1, so we can write $f = \sum_{i=0}^n a_i \big(\frac{1}{f}\big)^i$ in $B[\frac{1}{f}]$ for $a_i \in  A$. Clearing denominators shows that $f \in B$ satisfies a monic polynomial over $A$.


Observation 2 + killing the kernel shows:

**Theorem**: If $\phi:A \to B$ is a ring map that is universally closed on $\mathrm{Spec}$, then it is integral.