Never mind. Uncountably infinite is actually really easy to show.
by virtue of the fact that they cannot be obtained by the sum of anything larger, so they must be in any preimage.
By the same logic, for all
Now, either or not.
Let . It must be that because we have exhausted the sums of the smallest two elements, as well as the sum of a potentially 3rd smallest with that of the first two.
Now, let . Either or not.
This process can continue ad infinitum producing a binary sequence, each one representing a distinct member of . This maps bijectively with reals, and so my set is uncountable.
I realize I need a little more rigor to fully justify it, but the intuition is there, and that is good enough for now.