About the existence of a generic point on an irreducible closed subset of a prescheme

This is Proposition 2 on page 81 of Mumford's The Red Book of Varieties and Schemes:

Let $X$ be a prescheme, and $Z \subset X$ an irreducible closed subset. Then there is one and only one point $z \in Z$ such that $Z = \overline{\{ z \}}$.

Proof. Let $U \subset X$ be an open affine set such that $Z \cap U \neq \emptyset$. Then any point $z \in Z$ dense in $Z$ must be in $Z \cap U$; and a point $z \in Z \cap U$ whose closure contains $Z \cap U$ is also dense in $Z$. Therefore it suffices to prove the theorem for the closed subset $Z \cap U$. But by Prop. 1 of section 4 there is a unique $z \in Z \cap U$ dense in $Z \cap U$.

I have some questions about this proof.

1. Why does every dense point in $Z$ lie in $Z \cap U$?
2. Why must $z \in Z \cap U$ whose closure contains $Z \cap U$ be dense in $Z$?
3. Why is there a unique $z \in Z \cap U$ dense in $Z \cap U$? I can't find the proposition the author mentioned.

I think these are concerned with affineness, but I don't know the exact reason.

Thanks for everyone.

Solution 1:

1) A point $z$ dense in $Z$ must belong to any non-empty open subset of $Z$, so in particular to $Z\cap U$.
2) Any non-empty open subset $V\subset Z$ of an irreducible space $Z$ is dense: $\overline V=Z$.
So applying this to $V=Z\cap U$ we get $\overline {\{z\}}\supset \overline {Z\cap U}=\overline V=Z$, so that $\overline {\{z\}}=Z.$
3) Any non-empty open subset of an irreducible space $Z$ is irreducible, so $V=Z\cap U$ is irreducible. Hence $V\subset U$ is a closed irreducible subset of $U=\operatorname {Spec R}$.
By basic affine scheme theory $V=V(\mathfrak p)$ for some prime ideal $\mathfrak p\subset R$ and $[\mathfrak p]=z\in V$ is the required dense point of $V$.