For any $\varnothing\ne A\subset S$, let $\Sigma_A =\textstyle \sum_{x\in A}x^{-1}$. Note that for any $i_1,\dots,i_5\in\{1,2,\dots,10\}$, $0<\textstyle\sum_{j\le 5}\frac{1}{a_{i_j}} \le \sum_{j\le 5}j^{-1}\triangleq S$. That is, if $|A|=5$ then $\Sigma_A\in[0,S]$. Divide now $[0,S]$ into $\textstyle\binom{10}{5}-1$ equal pieces, each of size at most $S/\textstyle \bigl(\binom{10}{5}-1\bigr)<0.01$. By pigeonhole, there exists $D\ne D'$ so that $|\Sigma_D-\Sigma_{D'}|<0.01$. Finally, letting $E=D\cap D'$, it suffices to take $A=D\setminus E$ and $B=D'\setminus E$.