De Morgan's Laws proof 
My only problem with this proof is that they seem to be assuming that $(A \cup B)'$ is nonempty (similar idea for $A' \cap B')$. Is that because this holds trivially if $(A \cup B)'$ is empty? In other words, $A \cup B = U$ where $U$ is the universal set.

You are correct that this proof is assuming that $(A\cap B)'$ is nonempty. Luckily, we can quite easily check that it holds in that case, as it means that there are no elements not in both $A$ and $B$, and so $A'\cap B'=\emptyset$. Likewise for the other direction, if $A'\cap B'=\emptyset$ then there are no elements that are in neither $A$ nor $B$, so $(A\cup B)'=\emptyset$. The second part of the theorem is likewise easy to show.
