Good day, this problem is designed to use the Pigeonhole Principle. For a finite set A of integers, denote by $s(A)$ the sum of numbers in $A$. Let S be a subset of {$1,2,3,...,14, 15$} such that $s(B) \neq s(C)$ for any 2 disjoint subsets B,C of S. Show that $|S| \leq5$.
I proved that for $|S| \geq 7$ such subsets exist. Since the number of non-empty subsets are $2^7-2=126$, and the possible sums are $1\leq s(A)\leq 15+14+13+12+11+10=75$.
And then, according to the Pigeonhole Principle , putting $126$ sets into $75$ boxes of sums, there will be two sums such that $s(A)=s(B)$. But for $|S|=6$ this method doesn't work. It seems to me that if you prove for $|S|=6$, then for $|S| \leq 5$ it will become obvious.
Focusing on $|S| = 6$, as OP did $|S| \geq 7$.
First. see this question for how to deal with the case when $S$ is a subset of $\{1, 2, \ldots, \color{red}{14} \}$.
Naive PP gives us subset sums of 0 to 69, with only $2^6 = 64$ subsets, so it doesn't apply.
The refining idea is reduce the possible subset sums by considering the maximal difference between non-empty subsets of $S$, which is at most $10 + 11 + 12 + 13 + 14 = 60$, so there are at most 61 distinct possible non-empty-subset-sums. Since there are $2^6 - 1 = 63$ non-empty subsets, hence by PP, some two non-empty-subsets have the same sum.
(Read their solution if you need more details. Note: In a similar vein, other restrictions like "subsets with 2 or 3 elements" could work for similar problems.)
Back to this problem: When we apply this idea to the case when $S$ is a subset of $\{1, 2, \ldots, 15 \}$, if the maximal difference is 61 or lower, then we can apply PP per the above. However, since $11+12+13+14+15 = 65$, we have to take care of certain cases.
Consider what the 5 largest values of $S$ are, in order to get a maximal difference of 62 or higher. Show that they must be one of (I believe I got all cases, if not please let me know.)
For each case, we can easily find a conflicting subset sum from these 5 elements alone, by focusing on terms that differ by 1 or 2. Hence we are done.
With some tedious checking, $S=\{6, 9, 11, 12,13 \}$ has distinct subset sums. So we cannot strengthen the result further.