Let $L_1$ be regular and $L_2$ be non-regular and $L_1 \cap L_2 = \emptyset$. Assume $L_1 \cup L_2 $ is regular. Now, we have a DFA for $L_1 \cup L_2$ and also for $L_1$. So, to check if a string belongs to $L_2$, we have to make a DFA for $\bar{L_1} \cap (L_1 \cup L_2)$ which equals $L_2.$ $\bar{L_1}$ is regular as regular languages are closed under complement. And intersection of two regular languages is regular as regular languages are closed under intersection too. Thus we get $L_2$ as regular which is not possible. So, $L_1 \cup L_2$ cannot be regular.