Why is empty set an open set?
It's vacuously true. All points $\in\emptyset$ are interior points, have a blue-eyed pet unicorn, and live in Surrey.
Here's another perspective:
Suppose the empty set is not considered open.
Consider the sine function $\sin\colon \Bbb R \to \Bbb R$.
Then $(3\,.\,.\,4)$ is open (presumably), but $\sin^{-1}(3\,.\,.\,4)=\varnothing$ is not open, so the sine function is not continuous.
We don't want that!
Another thing: one of the key properties of open sets is that the intersection of any two open sets is open. If the empty set were not considered open, then that wouldn't be true anymore.
My answer would have nothing to compare with @HagenvonEitzen's wit and to the point, but here's a way to think of it:
The complement of an empty set is the whole set, which of course contains everything including all limit points. Hence the whole set is closed, and therefore it compliment, empty set is open.
Hagen's argument can be made to show empty set is closed and whole set is open. Because all points in empty sets are limit points, so empty set is closed. So its compliment, whole set is open.
Another (better) way to think of this:
A topological space generalizes the concept of a metric space. With this view, a function is continuous iff the inverse image of an open set is open. Since a function that maps the entire space onto a single point is always continuous, the empty set is open. Take an open set which does not contain the single point. Its inverse image is the empty set.
Above is a proof for the definition, however, empty set is open by the definition of a topology. But it's good to know why things are defined the way they are.
A final note: It should be "the" empty set, since it is unique.