Your conjecture is correct.
By the following argument, the increase in the number of coverable subsets when one more point is added to $n$ points is $\frac12\left(n^2+n+2\right)$, and the claim then follows by induction.
The argument consists of two parts: Counting the increase in the number of coverable subsets when an arbitrarily distant point is added, and showing that the number of coverable subsets doesn't change if points are moved.
Given $n$ points, we can choose the additional point far enough away that all the existing coverable subsets remain coverable. Thus, in this case the increase in the number of coverable subsets is given by the number of coverable subsets containing the new point. Since the new point can be chosen arbitrarily far away, the disks are effectively half-planes, so we're looking for the number of subsets including the new point coverable by a half-plane. Since these subsets correspond bijectively to partitions of the $n+1$ points into two subsets by a line, we need the number of such partitions. This is known to be
$\binom n0+\binom n1+\binom n2=1+n+\frac{n(n-1)}2=\frac{n^2+2+n}2\;.$
It remains to be shown that the number of coverable subsets doesn't change if points are moved. The number of coverable subsets could only change if a point is moved across a circle defined by three other points, since at all other positions existing disks can be adjusted to accommodate the movement; it takes three points to create an obstruction to such adjustments.
So consider which subsets become coverable or uncoverable when a point $x$ moves across a circle defined by three points $p$, $q$, $r$. Let's say $x$ crosses the arc of the circle that lies between $p$ and $q$, and let $A$ be the set of points inside the circle. Then the set $A\cup\{r,x\}$ can be covered if $x$ is inside the circle, but not if it is outside, since $p$ and $q$ are in the way. On the other hand, the set $A\cup\{p,q\}$ can be covered if $x$ is outside the circle, but not if it is inside, since excluding $x$ forces the circle to include $r$. These two changes cancel, and the total number of coverable subsets remains invariant.
Considering that the change happens when four points are on a circle and involves the two subsets formed by the points inside the circle with the two pairs of opposing points on the circle, I suspect that there's a nicer way of expressing this that's symmetric with respect to the four points, but I don't have the time to think about that right now – perhaps someone else can add it.