I think that this may answer your question.
Let $A$ be an arbitrary set.
Suppose that $A\cap \varnothing\neq \varnothing$ (suppose they are not disjoint).
Then there exists $x\in \varnothing$ such that $x\notin A$, which is a contradiction since the empty set contains no elements by definition.
Now, suppose that $\varnothing \nsubseteq A$. Again, we can see that this cannot be true since this would imply that there exists some element in $\varnothing$ which is not an element of $A$.
So in particular, the empty set is a subset of every set, and also that the empty set is disjoint with every set. Almost every time you prove something regarding the empty set, the best route to go is usually to use a proof by contradiction, since working with the empty set is almost always counter intuitive.