Disjoint unions of limited cardinality: is the following conjecture/proof correct? Is the following conjecture about collections of sets correct? And if so, does the proof work?
Conjecture. Let $\kappa$ denote an infinite cardinal number, and let $\mathcal{K}$ denote a collection of sets that is closed with respect to the following.


*

*Disjoint unions of cardinality less than or equal to $\kappa$.

*Complements (with respect to some set that is larger than or equal to $\bigcup \mathcal{K}$).

*Unions and/or intersections of cardinality strictly less than $\kappa$ (not necessarily disjoint).


Then $\mathcal{K}$ is closed with respect to unions and intersections of cardinality less than or equal to $\kappa$ (not necessarily disjoint).
Proof. Let $\mathcal{J} \subseteq \mathcal{K}$ denote a subcollection of cardinality less than or equal to $\kappa$. It will be shown that $\bigcup \mathcal{J} \in \mathcal{K}$.
Let $\lambda$ denote the least ordinal such that $|\lambda| = \kappa$, and choose a bijection $j : \lambda \rightarrow \mathcal{J}$. Define a function $k : \lambda \rightarrow \mathcal{K}$ as follows.
$$k(\beta) = \left( \bigcup_{\alpha<\beta} j(\alpha)\right)^c \cap j(\beta).$$
We will show that for all $\beta \in \lambda$ it holds that $k(\beta) \in \mathcal{K}$.

Fix $\beta \in \lambda$. Since $\lambda$ is the least ordinal with
  cardinality $\kappa$ and $\beta \in \lambda$, the union in the above
  expression has cardinality strictly less than $\kappa$, and is thus an
  element of $\mathcal{K}$. Thus its complement is an element of
  $\mathcal{\kappa}$. But since $\kappa$ is infinite, it follows that
  $\mathcal{K}$ is closed with respect to finite intersections, thus the
  above expression is an element of $\mathcal{K}$. So $k(\beta) \in$
  $\mathcal{K}$.

We conclude that for all $\beta \in \lambda$ it holds that $k(\beta) \in \mathcal{K}$. Furthermore, it can be shown that $$\bigcup_{\beta \in \lambda}k(\beta) = \bigcup_{\beta \in \lambda}j(\beta).$$
Noting that the expression on the left is a disjoint union of elements of $\mathcal{K}$ and has cardinality $\kappa$, we conclude that it is an element of $\mathcal{K}$. Thus the expression on the right is an element of $\mathcal{K}$. But the expression on the right equals $\bigcup \mathcal{J}.$ Therefore, $\bigcup \mathcal{J} \in \mathcal{K}$, as required.
 A: I think the conditions can be made stronger:
Theorem. Assume the class of sets $\mathcal K$ has these properties, where $\kappa$ is any cardinal:


*

*If $|I|\le \kappa$ and $A_i\in \mathcal K$ for $i\in I$ and $A_i\cap A_j=\emptyset$ if $i\ne j$, then $\bigcup_{i\in I}A_i\in\mathcal K$.

*If $A,B\in\mathcal K$ then $A\setminus B\in \mathcal K$.


Claim: If $|I|\le \kappa$ and $A_i\in \mathcal K$ for $i\in I$, then $\bigcup_{i\in I}A_i\in\mathcal K$.
Proof:
Let 
$$\mathcal L=\biggl\{\beta\in\mathrm{On}\biggm| |\beta|\le \kappa\text{ and }\exists f\colon\beta\to\mathcal K\text{ such that }\bigcup_{\alpha<\beta}f(\alpha)\notin\mathcal K\biggr\}.$$ 
Assume  $\mathcal L\ne\emptyset$ and let $\lambda=\min\mathcal L$.
Fix some $j\colon\lambda\to\mathcal K$ with $$S:=\bigcup_{\alpha<\lambda}j(\alpha)\notin\mathcal K.$$
For $s\in S$ let $i(s)=\min\{\alpha\in\lambda\mid s\in j(\alpha)\}$.
Then for $\alpha\in\lambda$,
$$ i^{-1}(\alpha)=j(\alpha)\setminus\bigcup_{\beta<\alpha}j(\beta).$$
Since $\alpha<\lambda$, we have $\alpha\notin\mathcal  L$ and $|\alpha|\le\kappa$, hence $\bigcup_{\beta<\alpha}j(\beta)\in\mathcal K$ and hence by property (2) also $i^{-1}(\alpha)\in\mathcal K$.
Now 
$$ S=\dot\bigcup_{\alpha\in\lambda}i^{-1}(\alpha)$$
shows that $S$ can be written as a disjoint union of $|\lambda|$ elements of $\mathcal K$. Since $|\lambda|\le\kappa$ and $S\notin \mathcal K$, we obtain a contradiction with property (1).
Therefore $\mathcal L=\emptyset$.
Final conclusion: If $I$ with $|I|\le\kappa$ is given, then there exists an ordinal $\beta$ and a bijection $g\colon\beta\to I$. With $f\colon\beta\to\mathcal K$, $\alpha\mapsto A_{g(\alpha)}$ we find that
$$\bigcup_{i\in I} A_i=\bigcup_{\alpha<\beta}f(\alpha)\in\mathcal K$$
because $|\beta|=|I|\le\kappa$ and $\mathcal L=\emptyset$. $_\square$
