Your boldface statement is false in general. Consider
$$A=F[X_1, \dots, X_n, ...., Y_1, \dots, Y_n, ....]/(X_n^2, X_nY_n)_n$$
over a field $F$. Denote by $x_n, y_n$ the images of $X_n, Y_n$ in $A$. Let $\mathfrak p$ be the ideal generated by the $x_n$'s. As $A/\mathfrak p$ is $F[y_1, \dots, y_n, ....]$ which is integral, $\mathfrak p$ is a prime ideal, generated by nilpotent elements, thus is the minimal prime ideal of $A$. As every $x_ny_n=0$, $\mathfrak pA_{\mathfrak p}=0$, hence $A_{\mathfrak p}$ is reduced.
If there were a reduced open neighborhood $U$ of $\mathfrak p$, then $U$ contains a non-empty reduced principal open subset $D(f)$. So $A_f$ is reduced and $\mathfrak p A_f=0$. Therefore for all $n\ge 1$, there exists $r_n\ge 1$ such that $f^{r_n}x_n=0$. Now it is easy to check that $f\in y_nA+\mathfrak p$ and $\cap_n (y_nA +\mathfrak p)=\mathfrak p$. Thus $f$ is nilpotent and $D(f)=\emptyset$. Contradiction.
Best Answer
Suppose that $U= \operatorname{Spec}A$ is an affine open of $X$ containing $x_0.$ Note that $U$ also contains $x$: if $x\not\in U,$ then we have $$ \overline{\{x\}} = (X\setminus U)\cap\left(\bigcap_{C\textrm{ closed}, C\ni x}C\right). $$ However, this implies $x_0\in X\setminus U.$
Now, we may suppose that $X = \operatorname{Spec}A.$ Suppose $\mathfrak{p}$ (respectively, $\mathfrak{p}_0$) is the prime of $A$ corresponding to $x$ (respectively, $x_0$). What does it mean for $\mathfrak{p}_0\in\overline{\{\mathfrak{p}\}}$? Well, since $\overline{\{\mathfrak{p}\}} = V(\mathfrak{p}),$ this means that $\mathfrak{p}\subseteq\mathfrak{p}_0.$ Remember that $\mathcal{O}_{X,\mathfrak{p}}\cong A_{\mathfrak{p}},$ and similarly for $\mathfrak{p}_0.$ However, if $\mathfrak{p}\subseteq\mathfrak{p}_0,$ then $A_{\mathfrak{p}}$ is indeed a further localization of $A_{\mathfrak{p}_0}$: you simply localize at $A_{\mathfrak{p}_0}\setminus\mathfrak{p}A_{\mathfrak{p}_0}$.