How can we argue $\nabla\cdot v = 0$ when ${\mathscr r} \ne 0$ on this vector function? I am dealing with the vector field:
$$v = \dfrac{\hat{\mathscr r} }{{\mathscr r}^2}$$
And I am studying its divergence. If we compute it we get:
$$\nabla\cdot\left(\dfrac{\hat{\mathscr r} }{{\mathscr r}^2}\right) = 0, \qquad {\mathscr r} \ne 0.$$
I understand we are dealing with a delta function, which explains why we get $\nabla\cdot v = 0$ everywhere but in the origin, where it blows up.
But the fact that $\nabla\cdot v = 0$ does not make sense to me looking at the graph of the vector function:

Where we can see how the vector field spreads out. The only reasoning I see is that at the origin, the vector field spreads out so much that once we look out of it the field cannot spread any more, thus we get zero divergence. But I insist, this cannot be seen in the plot of the function
 A: Despite the name, divergence does not say whether the field diverges in the meaning "spreads". A vector field can have zero divergence and be spreading, or it can have non-zero divergence and not be spreading. An example of the former, you have in your post. An example of the latter is $\vec{F} = x \hat{x}$ where $\hat{x}$ is the unit vector in the $x$ direction.
A: Your intuition about what divergence means is incorrect. Divergence measures how much field enters vs. leaves a very small volume about a particular point. At all points in this diagram except for the very center, if you were to zoom in enough, you would find a sufficiently small volume where you would see the following pattern: on the side of the volume closer to the center, the field would be slightly larger, but slightly more spread apart, while the field on the other side is smaller, but more uniform in direction. The effect of the spreading-out on the near side causes cancellation that reduces the amount of field poking perpendicularly through the surface on that side, so the field entering and leaving the volume are the same. Hence, there is no divergence  at points other than the center (where, no matter how small the volume, more field is leaving the volume than is entering it). At the center, in fact, the more you shrink the volume, the stronger the field is on the border, so the more field is leaving the volume; this goes to infinity, which means the divergence is also infinity.
Important note: The first version of this answer had an incomplete intuitive definition of divergence. This version has been fixed. 
