For $n=7$, we can take $2^2+2^2$ to write $7+1$. So suppose $n\geq 11$. Now we use the following lemma: Lemma: $x^2+y^2\equiv 1 \mod p$ has $p-(-1)^{\frac{p-1}{2}}$ solutions in $\mathbb{F}_p$. Proof: It is well known, that $x^2\equiv a \mod p$ has $\left(\frac{a}{p}\right)+1$ solutions in $\mathbb{F}_p$ where $\left(\frac{a}{p}\right)$ denotes the Legendre symbol. Then fix one variable and it follows that $x^2+y^2\equiv 1 \mod p$ has $$\sum_{k=0}^{p-1}\left(\frac{1-k^2}{p}\right)=p + (-1)\sum_{k=0}^{p-1}\left(\frac{c^2-1}{p}\right)=p+(-1)\cdot(-1)^{\frac{p-1}{2}}=p-(-1)^{\frac{p-1}{2}} \, $$solutions. $\blacksquare$ So $x^2+y^2\equiv 1 \mod p$ has at least $p-1$ solutions. Moreover, we can assume that $x, y \leq \frac{p-1}{2}$ since $x^2 \equiv (p-x)^2 \mod p$. Now we also don't consider the $4$ trivial solutions, so assume that none of $x,y$ is $0 \mod p$. Moreover, since $2\cdot\left(\frac{p-1}{2}\right)^2 = \frac{p^2-2p+1}{2}\equiv 2^{-1} \equiv 1 \mod p$ yields a contradiction, we must have $2 \leq x^2 +y^2 \leq \left(\frac{p-3}{2}\right)^2+\left(\frac{p-1}{2}\right)^2$. Especially, since we assumed that $p\geq 11$ (we have at least $4$ non-trivial solutions among $x,y\leq \frac{p-1}{2}$), we must have a solution to $x^2+y^2\equiv 1 \mod p$ s.t. $x^2+y^2\leq 2\cdot\left(\frac{p-3}{2}\right)^2$, but note that $2\cdot\left(\frac{p-3}{2}\right)^2 \leq \frac{(p-5)p}{2}+1$ for $p\geq 7$. Thus there exists a number among $S$, that can be written in the form $x^2+y^2$. This completes the proof. $\square$