The closure of $\overline{\{x\}}$ being irreducible and relating the generic point to its associated irreducible scheme 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 $\overline{\{\xi \}}=U$? 
 A: You already have two good answers; I just want to collect some general topological results with which one should become fluent when studying algebraic geometry.
A non-empty topological space $X$ is irreducible if the following equivalent conditions hold:


*

*If $Y_1, Y_2$ are closed subsets of $X$ and $X = Y_1 \cup Y_2$, then $Y_1 = X$ or $Y_2 = X$.

*If $U_1, U_2$ are non-empty open subsets of $X$ then $U_1 \cap U_2 \neq \emptyset$.

*Any non-empty open subset of $X$ is dense in $X$.


Proving that these are equivalent will probably be straightforward. Whether or not $X$ is itself irreducible, if $Y \subset X$ is irreducible in the subspace topology then so is its closure $\overline Y$ in $X$. In your notation, $\{x\}$ is certainly irreducible.
A: $1)$ if $\overline{\{x\}}$ was reducible, you would have a non-trivial decomposition $\overline{\{x\}}=F\cup F'$ in a union of two closed subsets. The complementary of, says, $F$ give then a non dense open subset.
$2)$ $U$ can certainly by equal to $\{\xi \}$, but if $U$ is a subscheme of $X$, it refers usually to closed subschemes (just as subvarieties).
A: 1) Since it implies that the intersection of two open subsets is not empty.
2) consider $X=U=\text{Spec} \; \mathbb{Z}$ with its generic point $\{ (0) \} \not = U$ while $\overline{\{ (0) \}}=V(0)=U.$
