Let $A$ be a valuation ring with maximal ideal $\mathfrak{m}$ which is the union of the non-maximal prime ideals of $A$. One can show that for $X=\text{Spec }A$, the open subscheme $X\setminus\{\mathfrak{m}\}$ has no closed points.
I'm looking for an explicit example of such a valuation ring.
Here are some facts which I have been trying to use. Let $\Gamma$ be the value group of $A$. We say a subgroup $\Delta\leq\Gamma$ is isolated if whenever $0_G\leq\gamma\leq\delta\in\Delta$, $\gamma\in\Delta$ as well. Let the $v$ denote the valuation on $A$.
- The isolated subgroups of $\Gamma$ are in 1-to-1, order reversing correspondence with the prime ideals of $A$. More precisely, given an isolated subgroup $\Delta$, the corresponding prime ideal of $A$ is the set of $a\in A$ with $v(a)\notin\Delta$.
- There can be no maximal $\mathfrak{p}\subsetneq\mathfrak{m}$, since otherwise $\mathfrak{m}$ would not be the union of the prime ideals it properly contains (then $\mathfrak{p}$ would be a closed point of $X\setminus\{\mathfrak{m}\}$. So there must be an infinite ascending chain of non-maximal primes, and thus an infinite descending chain of non-zero isolated subgroups (i.e. the isolated subgroups cannot be well-ordered).
- Let $k$ be a field. For any totally ordered abelian group $G$ the $k$-algebra $k[G]$ is a domain. We can define a function $v:k[G]\to G\cup\{\infty\}$ which takes in any nonzero $\sum a_{\gamma}\gamma$ and returns the least $\gamma$ with $a_{\gamma}$ nonzero, where here $\gamma$ ranges over some finite set of elements of $G$ (we define $v(0)=\infty$). Then $v$ can be extended to a valuation on $\text{Frac}(k[G])$. The corresponding valuation ring $R$ will have value group $G$.
- So if $G$ is a group with an infinite descending chain of non-zero isolated subgroups whose intersection is $\{0_G\}$, the maximal ideal of $R$ will be the set of elements of $\text{Frac}(k[G])$ with positive valuation. If $x\in R$ is such an element then $v(x)>0$ lies outside $\{0_G\}$ and so $v(x)$ is not in some non-zero isolated subgroup $\Delta$, and therefore the corresponding prime ideal of $R$ will contain $x$. This shows the maximal ideal of $R$ is the union of the non-maximal primes in $R$.
Therefore it suffices, for example, to find such a group $G$. Quite frankly I cannot even begin to imagine what such a beast would look like.
Okay, I was being really silly and getting my orders backwards when I tried to come up with examples of such groups $G$.
This is essentially equivalent to the example given by the link in the comment by math54321, except that uses (a group isomorphic to) $\mathbb{Z}^{\oplus\mathbb{N}}$.
Consider $\mathbb{Z}^{\mathbb{N}}$ with the lexicographical ordering. For any $n\in\mathbb{N}$, the elements $a=(a_0,a_1,\dots)$ of $\mathbb{Z}^{\mathbb{N}}$ satisfying $a_m=0$ for all $m<n$ form an isolated subgroup: if $0\leq b\leq a$ and $a_m=0$ for all $m<n$ then the first nonzero entry of $b$ cannot be negative and we can show by inducting on $m$ that for $m<n$, $0\leq b_m\leq a_m=0$ and therefore $b$ also lies in this group. Furthermore, the $0$ subgroup is the intersection of these subgroups over all $n$, which form a descending chain. (This is enough to perform the argument from the last bullet point at the end of the question.)