If $X$ is a scheme and $Z$ an irreducible closed subset, then $\exists\,\xi\in Z$ with $Z=\overline{\{\xi\}}$
Here is my attempt:
Let $X=\bigcup_i U_i$ be an open affine cover. Since $Z$ is irreducible, then $Z\cap U_i$ is a closed irreducible subset of the affine scheme $U_i$. This means there is a prime ideal $\mathfrak{p}_i\in U_i$ such that $Z\cap U_i=V_i(\mathfrak{p}_i)$ (where $V_i(I):=\{\mathfrak{p}\in U_i\mid \mathfrak{p}\supset I\}$)
On the other hand, $V_i(\mathfrak{p}_i)=\overline{\{\mathfrak{p}_i\}}^{U_i}$ (closure in $U_i$) or, equivalently, $V_i(\mathfrak{p}_i)=\overline{\{\mathfrak{p}_i\}}\cap U_i$. Therefore:
$$Z=\bigcup_iZ\cap U_i=\bigcup_i\overline{\{\mathfrak{p}_i\}}\cap U_i\subset\bigcup_i\overline{\{\mathfrak{p}_i\}}$$
Conversely, since $\mathfrak{p}_i\in Z$, then $\overline{\{\mathfrak{p}_i\}}\subset Z$ for all $i$, so $\bigcup_i\overline{\{\mathfrak{p}_i\}}\subset Z$ and:
$$Z=\bigcup_i\overline{\{\mathfrak{p}_i\}}$$
By irreducibility of $Z$, we have $Z=\overline{\{\mathfrak{p}_i\}}$ for some $i$. $_\blacksquare$
Is this correct?
I'm not sure about the conclusion since the union may be infinite, which made me suspicious.
Best Answer
I'll write down how I fixed the final argument using KReiser's suggestion.
In case $X$ is affine, $X$ is an affine open cover of itself, so $Z=V(\mathfrak{p})=\overline{\{\mathfrak{p}\}}$ for some $\mathfrak{p}\in X$ (notice that here $V(\mathfrak{p})$ is merely a set, not a scheme). Such $\mathfrak{p}$ is unique because $V(\mathfrak{p})=V(\mathfrak{p}')\Leftrightarrow \mathfrak{p}=\mathfrak{p}'$, since $\mathfrak{p},\mathfrak{p}'$ are prime. So we have proven:
Lemma: every irreducible closed subset of an affine scheme has a unique generic point.
Now suppose $X$ is any scheme. For simplicity, assume $Z\cap U_i\neq\emptyset$ for every affine open $U_i$ in the cover.
We already know that $Z\cap U_i=\overline{\{\mathfrak{p}_i\}}\cap U_i$ for some $\mathfrak{p}_i\in U_i$.
Claim: $\mathfrak{p}_i\in U_i\cap U_j$ for all $i,j$.
Proof: Suppose $\mathfrak{p}_i\notin U_j$. Then $\mathfrak{p}_i$ is in $U_i\setminus U_j$, which is closed in $U_i$, therefore $\overline{\{\mathfrak{p}_i\}}\cap U_i\subset U_i\setminus U_j$. Consequently $Z\cap U_i\cap U_j=\overline{\{\mathfrak{p}_i\}}\cap U_i\cap U_j=\emptyset$. On the other hand, $Z$ is irreducible and $Z\cap U_i$ and $Z\cap U_j$ are nonempty open sets of $Z$, so their intersection $Z\cap U_i\cap U_j$ is nonempty (contradiction). $\blacksquare$
Now take an affine open set $V$ such that $\mathfrak{p}_i\in V\subset U_i\cap U_j$. Notice that: $$\overline{\{\mathfrak{p}_i\}}\cap V=(Z\cap U_i\cap U_j)\cap V=\overline{\{\mathfrak{p}_j\}}\cap V$$
Since $V$ is affine, $\mathfrak{p}_i=\mathfrak{p}_j$ by the lemma. We might as well write $\mathfrak{p}=\mathfrak{p}_i$ for all $i$, so
$$Z=\bigcup_i\overline{\{\mathfrak{p}_i\}}=\overline{\{\mathfrak{p}\}}$$
Finally, if $\exists\,\mathfrak{p}'$ with $Z=\overline{\{\mathfrak{p}'\}}$, then $\overline{\{\mathfrak{p}'\}}\cap U_i=\overline{\{\mathfrak{p}_i\}}\cap U_i$ for any $i$. Since $U_i$ is affine, by the lemma, $\mathfrak{p}'=\mathfrak{p}_i=\mathfrak{p}\,\,_\blacksquare$