Let $f: X \to Y$ be a morphism of schemes. Let $\Delta : X \to X \times_Y X$ denote the diagonal morphism. Take $U = \textrm{Spec } A$ and $V = \textrm{Spec } B$ be affine open subsets of $Y$ and $X$ respectively such that $f(V) \subseteq U$.
$\Delta (X) \cap (V \times_U V)$ is a closed subscheme of $V \times_U V =\textrm{Spec } B \otimes_A B$ defined by the kernel of the diagonal homomorphism $B \otimes _A B \to B$
My attempt:
Let $\delta$ denote the diagonal homomorphism of rings. Then it is clearly surjective so $B \cong (B \otimes B) /\ker \delta$. Moreover, we have $B \to B \otimes B \to B$ factoring through $\delta$ so this gives a closed immersion of schemes $\textrm{Spec} ((B \otimes B) /\ker \delta) \to \textrm{Spec} (B \otimes B)$.
I now wish to identify the former with $\Delta (X) \cap (V \times_U V)$. $V \times V$ gives the prime ideals of $B \otimes B$ so it remains to see that the prime ideals containing the kernel of $\delta$ is precisely $\Delta (X)$. Here I am unsure how to proceed because $X$ is a general scheme, not necessarily affine. Even restricting to the affine subset $V$ and considering $\Delta (V)$, I can't quite see the result.
This is just abstract nonsense – nothing specific about schemes at all. We start with a commutative square of the form below, $$\require{AMScd} \begin{CD} V @>>> X \\ @VVV @VVV \\ U @>>> Y \end{CD}$$ where the horizontal arrows are monomorphisms. The claim is that the following is a pullback square: $$\begin{CD} V @>>> X \\ @VVV @VVV \\ V \times_U V @>>> X \times_Y X \end{CD}$$ Since $U \to Y$ and $V \to X$ are monomorphisms, $V \times_U V \to X \times_Y X$ is a monomorphism as well. Now the claim is clear: given a morphism $T \to X \times_Y X$ that factors through both $X \to X \times_Y X$ and $V \times_U V \to X \times_Y X$, it must also factor through $V \to X \times_Y X$.