First, given an uncountable index set $I$ and nonnegative numbers $d_i$ for $i\in I$, what do we even mean by the expression $\sum_{i\in I} d_i$? Well, we mean the following,
$ \sum_{i\in I} d_i = \sup_{F\subset I} \sum_{i\in F} d_i, $ where $F$ is finite (You can give a definition when the $d_i$ are no longer nonnegative, but then the sum isn't necessarily defined, i.e., doesn't converge to anything).
Now that we have this definition, we claim the following proposition.
Proposition. Given any uncountable collection $d_i>0$ of positive numbers indexed by $I$, the sum $\sum_{i\in I} d_i$ is infinite.
Proof. Let $E_0=\{i\in I\mid d_i>1\}$ and $E_n=\{i\in I\mid \frac{1}{n+1}. Then the $E_n$ are disjoint and their union is $I$. As a result, there is some $k$ for which $E_k$ is infinite (otherwise $I$ would be a countable union of finite sets and therefore countable). Then we can choose arbitrarily large finite subsets $F\subset E_k\subset I$ and for these we have
$ \sum_{i\in F} d_i \ge \sum_{i\in F} \frac{1}{k+1} = \frac{\vert F\vert}{k+1} $ Since $k$ is fixed, the right-hand side can become arbitrarily large, and hence the supremum over all finite subsets of $I$ is infinite. #
Now, an application of this proposition is that if the sum over every finite subset of $I$ is uniformly bounded (say by $\Vert v\Vert^2$), then the sum over $I$ is finite and less than or equal to this uniform bound. Hence, the set $\{i\in I\mid d_i>0\}$ is at most countable.