 Solution

Take $S$ to be the primes in the interval and $Q=\max (2,\lfloor \sqrt{N}\rfloor )$. Every such prime exceeds $M>N\ge Q$ apart from harmless bounded small cases, so it avoids zero modulo each prime up to $Q$. The sifted-interval bound with $H=N$ and $q=1$ givesSince $Q^2=O(N)$ and $\log (2Q)\asymp \log N$, the implied constant is absolute. The choice of strict or inclusive endpoint in the prime-counting convention changes at most two terms, absorbed by the bound for $N\ge 2$.