Looking at this question, I saw user Zhen Lin's comment that when $X$ is a separated scheme over an affine scheme $S$, and $U$ and $V$ are affine open subsets of $X$, we can identify $U \cap V$ with $(U \times_S V) \times_{X \times_S X} X$. I've been able to verify that $U \cap V$ has the universal property of $(U \times_S V) \times_{X \times_S X} X$, but I'm not sure why this tells me that $U \cap V$ is affine?
I know that $U \times_S V$ is affine since $U, V$, and $S$ are affine, but $X$ and $X \times_S X$ might not be affine, so I assume somehow the separatedness of $X$ over $S$ guarantees that $(U \times_S V) \times_{X \times_S X} X$ is affine?
To be clear, I understand how to prove that $U \cap V$ is affine using that $\delta$ is a closed immersion, and I know that this should then imply that $(U \times_S V) \times_{X \times_S X} X$ is affine, but the comment on the linked post made me think there is an alternative proof using properties of separated morphisms to see that $(U \times_S V) \times_{X \times_S X} X$ is affine directly, and I'm asking what that proof would be?