An easy way to satisfy A is to add a constant c to each member of the set, where c is equal to or larger than the largest number in the set.
If we have the set [x, y and z] where z is largest, we can add z to every element. Then the new set [x+z, y+z, 2z]. The sum of the two smallest elements is 2z+x+y, which is of course larger than 2z.
(Edit: we can actually use a constant that is smaller than z, because condition A does not require the single element to be smaller than the sum of any two elements; it just can't be larger. That means it can still be equal! If we use z-1, then the new set would be [x+z-1, y+z-1, 2z-1] and the sum of the two smallest elements would be 2z+x+y-2. If x+y = 1, then the sum is equal to the largest element and the condition is still satisfied.)
An easy way to satisfy B is to make each subsequent number the sum of all previous numbers plus a constant > 0. Then your set will be:
[a, a+b, 2a+2b, 4a+4b, 8a+8b...]
If you use a = 1 and b = 1, then you just get powers of 2.
Because of the way that the numbers grow, it is impossible to make two subsets equal. Suppose you have a subset S1 where the highest number is x. When you construct the second subset S2, if you use a number from the set that is larger than x, it is guaranteed to be larger than sum of values in S1. If you do not use a number larger than x, then the sum of values in S2 is guaranteed to be smaller.
Combining these two methods shouldn't cause any conflict. The constant c is already large enough that any equality in sums should still be impossible. That is, the constants guarantee that the sum of S1 will always be larger than the sum of S2 if |S1| > |S2|. If the two subsets have the same cardinality, then method B guarantees inequality (adding the same constant to either side changes nothing).
Assuming I haven't done something stupid in this thought process, the smallest version of my solution would be...
Well, for method B, we can use [0, 1, 2, 4, 8, 16, 32, 64]. We can start with 0 because we are going to add a constant anyways.
For this set with method A, we need to add 64. So then the final set is [64, 65, 66, 68, 72, 80, 96, 128].
That is the set that Drab mentions immediately after silverspawn's solution. There probably are smaller sets that take advantage of the fact that subsets are restricted to a maximum of 3 elements.
(Edit: we could actually use a constant of 63, as per my edit for method A above. So the set would be [63, 64, 65, 67, 71, 79, 95, 127].)
(Edit edit: no, that doesn't work. 63+64+65 = 127+65.)
I am not sure how pacovf found his "smallest version of silverspawn's solution".
(Edit edit: OK, this is more complicated than I'd thought...)