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Source. Lemma 4, printed p. 88 (PDF p. 4).

Statement. Fix $0<c<1$. Suppose a family of $r$ distinct integers at most $x$ has $r\ge x/(\log x)^c$, and at least $r/2$ of its members have a proper divisor $d$ such that all prime factors of $n/d>1$ exceed $d(\log x)^{10}$. Then, for all sufficiently large $x$, there is one integer $d$ corresponding in this sense to more than

$$\frac{x}{d(\log x)^5}$$

members of the family.

Full proof

For each qualifying member choose one such divisor, for example the least. Let $s_d$ count the members assigned to $d$. These assignments are finite, $1\le d\le x$, and $\sum_d s_d\ge r/2$.

If every $s_d$ were at most $x/[d(\log x)^5]$, then, writing $X=\log x$,

$$\frac{x}{2X^c}\le\frac r2 \le\sum_{d\le x}s_d \le\frac{x}{X^5}\sum_{d\le x}\frac1d \le\frac{x(1+X)}{X^5} <\frac{x}{2X^c}$$

for all sufficiently large $x$, a contradiction. Therefore one $s_d$ is strictly greater than the required threshold. Every assigned divisor retains the prime-gap property.

Source precision. The sentence before the source's summation must be read as existence of at least one witness divisor for each of the qualifying integers. A single divisor common to many members is the conclusion of the summation, not an assumption. Choosing one witness per integer makes the counting unambiguous.

Use. The upper proof of Theorem 2 applies this after Lemma 3 has discarded $o(r)$ exceptions.