Why do we need topological distinctiveness if the points in the set of a topological space are distinct to start with? According to the Wikipedia article on the Separation axiom:

Let X be a topological space.  Then two points x and y in X are topologically distinguishable if they do not have exactly the same neighbourhoods (or equivalently the same open neighbourhoods); that is, at least one of them has a neighbourhood that is not a neighbourhood of the other (or equivalently there is an open set that one point belongs to but the other point does not).

This seems to be just dressing up the basic idea that given any point in a topological space, some points are closer to it than others, while remaining distinct no matter how close other points are. But if this is so:
Why do we still need topological distinctiveness if the points in the set of a topological space are distinct to start with?
 A: It is not a property of the set that the topology is defined on, but the topology itself (which a priori doesn't know about the elements of the set).
Where such a concept is useful is in something like the following circumstance (as well as others):
We have two topological space $(X,\tau)$ and $(Y,\tau')$ with the same underlying set and so, as sets $X=Y$. We want to ask whether these two spaces are homeomorphic. That is, are they the 'same' topological space for all intents and purposes. Note that they can still be homeomorphic but have $\tau\neq\tau'$. One way to check this would be to show that all the points in $X$ are topologically distinguishable (for instance suppose that $\tau$ is the discrete topology), and none of the points of $Y$ are topologically distinguishable (for instance suppose that $\tau'$ is the indiscrete topology). If this is the case, we can conclude that $X$ and $Y$ are not homeomorphic spaces.
That is, the property 'all points are topologically indistinguishable' is a topological invariant which means that if $X$ and $Y$ are homeomorphic, and the property holds for $X$, then the property also holds for $Y$.
