With induction is simple: for $ n=1$ it's obvious. Denote $ \sup(S_{n}(d))=u$, then a number $ s>u-e$($ 0<e<u$ , $ e$ fixed) will have an element(suppose $ x_{n}$) $ d>\frac{1}{x_{n}}> \frac{u-e}{n}\rightarrow \left[\frac{1}{d}\right]<x_{n}\le \left[\frac{n}{u-e}\right]$, so it can take a finite number of values. By induction, for every $ x_{n}$ , $ S_{n-1}\left(d-\frac{1}{x_{n}}\right)$ will have a maximum element. So we will have a finite number of maximum elements, adding to each one it's $ \frac{1}{x_{n}}$ we can choose a maximum element for $ S_{n}(d)$.