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Source. Lemma 2, printed p. 87 (PDF p. 3).

Statement. The number of positive integers $n\le x$ divisible by $p^2$ for some prime $p>\log x$ is $o(x/\log x)$.

Full proof

The number is at most

$$\sum_{p>\log x}\left\lfloor\frac{x}{p^2}\right\rfloor \le x\sum_{p>\log x}\frac1{p^2}.$$

For large $Y$, split the primes above $Y$ into intervals $(2^jY,2^{j+1}Y]$, $j\ge0$. The prime number theorem implies the uniform upper estimate $\pi(t)\le C t/\log t$ for all sufficiently large $t$. Thus

$$\sum_{p>Y}\frac1{p^2} \le\sum_{j\ge0}\frac{\pi(2^{j+1}Y)}{(2^jY)^2} \le\frac{2C}{Y\log Y}\sum_{j\ge0}2^{-j} \ll\frac1{Y\log Y}.$$

Taking $Y=\log x$ gives $O(x/(\log x\log\log x))=o(x/\log x)$, as required. No independence of prime divisibility events is assumed.

Use. Lemma 3 can therefore take every prime factor exceeding $\log x$ with exponent one, after deleting the stated exceptional set.