Consider the nature of vacuously true statements.
Let $x \in \emptyset$. There is no such element so anything we say about is vacuously true. Is $x$ a green dragon? Yes. Since there are no $x \in \emptyset$ then every $x\in \emptyset$ (all zero of them) are green dragons. This is true because there aren't any $x \in \emptyset$ that aren't green dragons.
Suppose $r \in$ extended reals and $x\in \emptyset$. Is $r \le x$? Yes. Since there is not $x\in \emptyset$ that is vacuously true. $r$ is less than or equal to every element in $\emptyset$ because there aren't any elements in the $\emptyset$ were it is not true.
So the empty set is bounded above by $r$ and likewise for any $r\in$ extended reals that we choose. So what is the least upper bound. Well, as all extended reals are upper bounds that would be the same thing as asking what is the least extended real. And clearly that is $-\infty$.
So $\sup \emptyset = -\infty$.
If that's to breezy we have the definition of $\sup \emptyset = s$ is 1) it is an upper bound. and 2) the is no $r < s$ that is an upper bound.
Well is that true of $-\infty$?
1) if $x \in \emptyset$ then is $x \le -\infty$? Yes. That is vacuously true because there is no $x \in \emptyset$?
2) Is there any $r < -\infty$ that is an upper bound of $\emptyset$. Well, no, there isn't any $r < -\infty$ at all.