Firstly note that since every polygon is bounded we can find a positive integer $N$ such that $S$ is inscribable inside a $(N-1)\times (N-1)$ square (with sides parellel to the $x$ and $y$ axes). Therefore, for Nicole it suffices to find $A,B\in\mathbb{Z}$ such that $\forall i,j\in [1,N]$ we have $(A+i,B+j)=1$. To do this, pick $N^2$ distinct primes, and call them $p_{i,j}$, with $1\leq i,j\leq N$. Now, by chinese remainder theorem, there exists an integer $A$ such that $A\equiv -i\pmod{p_{i,j}}$ for all $i,j$ since the primes are all distinct, and similarly there exists $B$ such that $B\equiv -j\pmod{p_{i,j}}$ for all $i,j$. Now, since $p_{i,j}|(A+i,B+j)$, it follows that Nicole has a winning strategy.