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Write $d(c)=y-x$, $p_c=q+d(c)$ and $w(c)=2y-x$ for a cell $c=(x,y)$ of $S$. Since $d$ is injective on $S$, the $|S|$ primes $p_c$ are distinct. Put $K=\lfloor\alpha P\rfloor$.

Size of the primes. From $|x|,|y|\le\omega$ we get $|d(c)|\le2\omega$, so every $p_c$ and $P$ lie in $[q-2\omega,q+2\omega]$. The hypothesis $(4-\alpha)q\ge16\omega+1$ with $4-\alpha\le1$ gives $q\ge16\omega+1$. Since $3P$ is an integer and $3P\le\alpha P$, we have $K\ge3P$. Hence

$$K\ \ge\ 3q-6\omega\ \ge\ 2q+4\omega+1, \qquad K\ \le\ \alpha(q+2\omega)\ \le\ 4q-8\omega-1 .$$

The first inequality uses $q\ge10\omega+1$. The second is $(4-\alpha)q\ge(2\alpha+8)\omega+1$, which follows from the hypothesis because $2\alpha+8<16$.

The layers. By the Chinese remainder theorem choose $z$ with $z\equiv-x\pmod{p_c}$ for every $c=(x,y)$ in $S$. The multiples of $p_c$ are then exactly the integers

$$z+x+tp_c\;=\;z+tq+(1-t)x+ty\qquad(t\in\mathbb Z),$$

since $x+t(y-x)=(1-t)x+ty$. Call this the layer-$t$ multiple of $c$. With $|x|,|y|\le\omega$ its offset from $z+tq$ is at most $(|1-t|+|t|)\omega$ in absolute value. In particular:

• layer $0$ is $z+x\in[z-\omega,z+\omega]$;
• layer $1$ is $z+q+y\in[z+q-\omega,z+q+\omega]$;
• layer $2$ is $z+2q+w(c)\in[z+2q-3\omega,z+2q+3\omega]$;
• every layer $t\le-1$ is at most $z-q+3\omega$;
• every layer $t\ge3$ is at least $z+3q-5\omega$.

For the last two, the bound at $t=-1$ is $z-q+3\omega$ and at $t=3$ it is $z+3q-5\omega$. Moving from $t$ to $t-1$ or $t+1$ changes the position by $q$ and the offset bound by $2\omega<q$, so the bounds hold for all $t\le-1$ and all $t\ge3$.

The block. Let

$$L=\max\bigl(z-q+3\omega+1,\ z+2q+3\omega+1-K\bigr), \qquad J=[L,\,L+K-1].$$

We check $z-q+3\omega+1\le L\le z-\omega$ and $z+2q+3\omega\le L+K-1\le z+3q-5\omega-1$. The two lower bounds hold by the definition of $L$. For the two upper bounds, compare each term of the maximum with each upper limit: $z-q+3\omega+1\le z-\omega$ since $q\ge4\omega+1$; $z-q+3\omega+1\le z+3q-5\omega-K$ since $K\le4q-8\omega-1$; $z+2q+3\omega+1-K\le z-\omega$ since $K\ge2q+4\omega+1$; and $z+2q+3\omega+1-K\le z+3q-5\omega-K$ since $q\ge8\omega+1$.

Therefore $J$ contains every layer $0$, $1$ and $2$ multiple: they lie in $[z-\omega,z+2q+3\omega]\subseteq J$. It contains no multiple in a layer $t\le-1$ or $t\ge3$, since those lie below $L$ or above $L+K-1$. So the integers of $J$ divisible by some $p_c$ are exactly the three sets $z+X$, $z+q+Y$ and $z+2q+W$.

Counting once. The three sets lie in the disjoint intervals $[z-\omega,z+\omega]$, $[z+q-\omega,z+q+\omega]$ and $[z+2q-3\omega,z+2q+3\omega]$, because $q>4\omega$. Hence $J$ contains exactly $|X|+|Y|+|W|$ integers divisible by at least one $p_c$, each counted once; an integer of $J$ divisible by several of the primes is a single element of one of these sets.

Positivity. Replacing $z$ by $z+n\prod_{c\in S}p_c$ preserves every congruence and translates $J$ and all layers by the same amount. For $n$ large, $L\ge1$, so $J$ is a block of $K$ consecutive positive integers. This completes the proof.