Given is a set $S$ and a function $f$ that operates on it $f : \mathcal{P}(S) \to \mathcal{P}(S)$. The claim is that if $f$ only uses $\cap$ and $\cup$ in its definition, then it is order-preserving, i.e:
If $R \subset S$, then $f(R) \subset f(S)$ for all sets $S$ and $R$.
I am not looking for a formal proof, but rather just an intuition as to why this always holds.
Well, binary union and intersection are themselves monotonic, in that, for any sets $A_0$, $A_1$, $B_0$, and $B_1$ such that $A_0 \subset A_1$ and $B_0 \subset B_1$, $A_0 \cup B_0 \subset A_1 \cup B_1$, and $A_0 \cap B_0 \subset A_1 \cap B_1$. And, furthermore, compositions of monotonic functions are monotonic. So compositions of binary union and intersection should be monotonic.