Proof of the Reduced-to-Separated theorem

Question

I'm trying to understand Vakil's proof of the Reduced-to-Separated theorem:

Theorem: Two $S$-morphisms $\pi:U\to Z$ and $\pi':U\to Z$ from a reduced scheme to a separated $S$-scheme agreeing on a dense open subset of $U$ are the same.

Proof: Let $V$ be the locus where $\pi$ and $\pi'$ agree (we have just proved that this exists). It is a closed subscheme of $U$ (because $Z$ is separable) which contains a dense open set. But the only closed subscheme of a reduced scheme $U$ whose underlying open set is dense is all of $U$.

The sentence I've written in bold is the sentence I don't understand. Vakil doesn't say anything further, and I don't remember ever proving this fact (though I may have missed it somewhere). Can somebody help me see why this is true?

Answer 1

It is sufficient to show this for an affine scheme $U=Spec(R)$. If we have a closed subscheme $X=Spec(R/I)$, it is enough to show that $X$ doesn't contain some point in $U$, as then if $U'$ is an open dense subset of $X$, $\bar{U}=X\neq U$. Assume $X$ contains all points $p\in U$. Thus as ideals, $I\subset p$ $\forall p\in U$, thus $I$ is contained in the nilradical of $R$, so unless $I=0$ and $X=Spec(R)$, $U$ is nonreduced, a contradiction.