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 $\phi^{-1}(B^\ast) = A^\ast$.
Proof: Fix $a \in A$ with $\phi(a) \in B^\ast$. We must show $a \in A^\ast$. Fix any prime $\mathfrak{p}$ of $B$. Then the map $A' := A/\phi^{-1}(\mathfrak{p}) \to B/\mathfrak{p} = B'$ is also closed on $\mathrm{Spec}$. Since the generic point of $\mathrm{Spec}(A')$ is in the image, all points of $\mathrm{Spec}(A')$ are in the image by closedness. Then $a$ must be a unit on $A'$: otherwise $\phi(a)$ would be contained in a prime of $B'$. Varying over all $\mathfrak{p}$ shows that $a$ is a unit modulo $A \cap \Big(\cap_{\mathfrak{p}} \mathfrak{p}\Big) = A \cap \mathrm{Nil}(B) = \mathrm{Nil}(A)$, where I use injectivity for the last equality. This forces $a \in A^\ast$.
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.