Here is my short solution. Define $\mathbb{S}_k=\{x: a_k|x \text{ and } x\in\mathbb{S}\}$ for each natural number $k$ less than or equal to $n$. It is easy to show that any two such sets are pairwise disjoint. $|\mathbb{S}|=m$ so $|\mathbb{S}_k|=\left\lfloor\frac{m}{a_k}\right\rfloor$ So $\sum_{k=1}^{n}\left\lfloor\frac{m}{a_k}\right\rfloor=\sum_{k=1}^{n}|\mathbb{S}_k|\leq |\mathbb{S}|=m$ So $\sum_{k=1}^{n}\left(\left\lfloor\frac{m}{a_k}\right\rfloor+1\right)\leq m+n$ Now for each $k$, $\frac{m}{a_k}<\left\lfloor\frac{m}{a_k}\right\rfloor+1$ So $\sum_{k=1}^{n}\frac{m}{a_k}< m+n$ So $\sum_{k=1}^{n}\frac{1}{a_k}< \frac{m+n}{m}$ as required.