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Source. Equation (19) and the following argument, printed p. 89 (PDF p. 5).

Let $\mathcal N$ be the set of square-free integers

$$n=p_1\cdots p_k,\qquad k\ge2,\quad p_1=3,\quad p_2=5,\quad p_1<\cdots<p_k,$$

such that

$$p_i<\prod_{j<i}p_j\qquad(3\le i\le k). \tag{1}$$

Statement. Every finite subset $N$ of $\mathcal N$ satisfies $g_N(d)\le d$ for every integer $d\ge1$.

Full proof

Consider $s$ distinct members $n_1,\ldots,n_s$ whose pairwise gcd is $d$. If $s\le1$, then $s\le d$ immediately. Suppose $s\ge2$. Then $d$ is square-free and divisible by $15$, since all members contain the primes 3 and 5. Write $n_j=dv_j$. The $v_j$ are pairwise coprime; at most one of them is one.

If $v_j>1$, take the least prime factor $p_i$ of $n_j$ not dividing $d$. It is also the least prime factor of $v_j$. The index is at least three, and every earlier prime factor divides $d$. Hence

$$p_i<\prod_{h<i}p_h\le d$$

by (1). Assign this prime to $v_j$. Pairwise coprimality makes the assigned primes distinct. There are at most $\pi(d-1)$ such primes, so

$$s\le1+\pi(d-1)<d\qquad(d\ge15).$$

For the last inequality, among $1,\ldots,d-1$ at least 1 and 4 are not prime, so $\pi(d-1)\le d-3$. This proves the result in every case.

Source precision. The source says every cofactor $v_j$ has a prime smaller than $d$. A member $n_j=d$ instead gives $v_j=1$. There is at most one such member, and the additional one above is harmless. Singleton gcd families are treated before division by $d$, since their pairwise gcd condition alone does not imply $d\mid n_j$.

Use. The [[covering_systems/erdos_1968_problem_p_erdos_s_stein/theorem_2_lower_bound|lower construction for $F(x)$]] counts elements of this fixed infinite sequence. It does not assert that they admit a disjoint progression system.