You don't need to show that $(y,y) \in R$, you need to show that for every $S \subseteq X$, that $(S,S) \in R$.

What this amounts to, is showing that for every subset $S$ of $X$, that $S \subseteq S$.

Now to show THIS, we can take any element $y \in S$, and we must show that this implies $y \in S$. Since this is (OBVIOUSLY!) always true, $S \subseteq S$, that is $(S,S) \in R$, so $R$ is reflexive.