Are Boolean rings integral domains?
My assumption was no. The "product" in Boolean rings is the intersection $\cap$ of two sets from $\mathscr{P}(X)$. If Boolean rings we're integral domains, then for $A, B\in \mathscr{P}(X)$ I should have that $$A \cap B = \emptyset \implies A= \emptyset \text{ or } B = \emptyset.$$ This is not the case. If $A=\{1,2\}$ and $B=\{3,4\}$, then $A$ and $B$ are disjoint and $A\cap B=\emptyset$, but neither one of $A$ or $B$ is empty. Is my reasoning here correct or am I making some flaws?
It is not true that every Boolean ring is of the form $\mathcal{P}(X)$ for some set $X$, but your reasoning is at least correct for those rings.
More generally, in a Boolean ring $x^2=x$ for all $x$, so if the ring were integral, that would mean that $x\in \{0,1\}$ (since $x(x-1)=0$). So if there are more than two elements in the ring those two properties are incompatible.
This is actually a pretty strong opposition: a Boolean ring is a ring where every element is idempotent; but the existence of non-trivial idempotents corresponds to the ring not being connected, and being integral is stronger than being connected. So in some sense the more idempotents you have, the less integral you are. (If this paragraph does not make that much sense to you, you can ignore it.)