Complexity classes not closed under intersection and union
The class $AWPP$ is not known to be closed under union, though it is easy to show it is closed under intersection
For the class $A_0PP$ it's the other way around: it is closed under intersection, but not known to be closed under union.


These classes have nice machine model definitions, namely they are something akin to an abstract version of the quantum complexity classes $BQP$ and $QMA$, respectively. These classes, in turn, are the quantum analogues of $P$ and $NP$, respectively. Let me give the definition (The Zoo) of $AWPP$:

Definition$^{(\ast)}$ (AWPP) A language $L\subseteq \{0,1\}^\ast$ is in $AWPP$ when there is a Gap function $g\colon \{0,1\}^\ast\to \mathbb{Z}$ satisfying the following Soundness and Completeness properties for a polynomial $p(x)$ and a polynomial-time computable function $f(x)$:


Soundness If $x\not\in L$ then $0\leq g(x)\leq 2^{-p(x)}f(x)$
Completeness If $x\in L$ then $(1-2^{-p(x)})f(x)\leq g(x)\leq f(x)$


These soundness and completeness conditions are precisely what you get if you were to (1) trace through the definition of $BQP$ using only the Hadamard gate and classical gates and then (2) interpret the acceptance probabilities as "paths", and then (3) impose that there is a large gap in acceptance probabilities depending on whether $x\in L$ or not.

It is interesting that this is not known, because both $BQP$ and $QMA$ are known to be closed under intersection and union. As far as I know, there have been no publications exploring the closure properties of these classes; perhaps they are a bit niche, or perhaps I am simply ignorant.

Of course it is also unknown whether $AWPP$ and $A_0PP$ are closed under complement, because this would imply the other closures as well; the converse does not hold; they may be closed under union and intersection but not complement, like $NP$ probably is.

$^{(\ast)}$In this definition, the polynomial $p(x)$ can be replaced by a polynomial $p^\prime(n)$ which depends only on $n=|x|$ the length of the string; similarly the function $f(x)$ can also be replaced by such a polynomial $f^\prime(n)$. This may make certain things easier to prove.