If $x$ is a point in $X$ where $X$ is a scheme, we write $\overline{\{ x\}}$ for the closure of $x$ in $X$.
$\mathbf{Question \;1}$: I am a bit confused why $\overline{\{ x\}}$ is irreducible. According to some lecture notes, this scheme $\overline{\{ x\}}$ is irreducible since an open subset of $\overline{\{ x\}}$ that doesn't contain $x$ also doesn't contain any point of the closure of $x$ since the complement of an open set is closed. Therefore, every open subset of $\overline{\{ x\}}$ contains $x$, and is therefore dense in $\overline{\{x \}}$.
So how does every open subset of $\overline{\{x \}}$ being dense relate to $\overline{\{x \}}$ being irreducible?
$\mathbf{Question \;2}$: Suppose $X$ is a scheme and $U=\operatorname{Spec }A$ is a nonempty irreducible subset of $X$. Then $U$ has a unique generic point $\xi$ corresponding to the minimal prime of $A$. Why is it that $\{ \xi\}\not=U$, but instead $\overline{\{\xi \}}=U$?