The proof of Lemma 3.4 uses the constant $K=[\epsilon^{-3}]$ which should be your clue. After the definition of $K$ the author covers $(8^t,8^{t+\epsilon})$ (which contains at least $\epsilon 8^n/(2n)$ primes by the choice of $t$ as per previous paragraph and Lemma $3.1$) with $K$ intervals each having at most $O(\epsilon^2 8^n/(n))$ primes by Lemma $3.3$ (see the at most vs at least above)
In particular, if only one (or none) of those intervals contains at least $O(\epsilon^4 8^n/(n))$ primes (so the rest contain at most $o(\epsilon^4 8^n/(n))$ primes) and since there $K \approx \epsilon^{-3}$ such intervals, the total number of primes in $(8^t,8^{t+\epsilon})$ can be at most $$O(\epsilon^2 8^n/n)+\epsilon^{-3}o(\epsilon^4 8^n/n)=o(\epsilon 8^n/(2n))$$ and that is a contradiction with the above. So the claim follows - and actually one gets that at least of the order $\epsilon^{-1}=K^{1/3}$ such nonconsecutive $a,b,....$ exist
Edit to answer the question from comment: What I mean by $O(\epsilon^2 8^n/n)+\epsilon^{-3}o(\epsilon^4 8^n/n)=o(\epsilon 8^n/(2n))$ is that in the paper Lemma $3.3$ bounds the number of primes in any one of those smaller intervals by $C\epsilon^2 8^n/n$ where $C=C_{\epsilon_0}$ is valid for all $\epsilon <\epsilon_0$ and all $x$ (hence all $n,t$ intervals) large enough independent of $\epsilon_0$; so now if for a positive constant to be chosen later $D$, there is only at most one interval of those $K=[\epsilon^{-3}]$ that has at least $D\epsilon^4 8^n/(n)$ primes, then the total number of primes in the large interval is at most $C\epsilon^2 8^n/n+DK\epsilon^4 8^n/n \le D_1\epsilon 8^n/(2n)$ for a given $D_1$ depending only on $D, \epsilon_0$ and going to zero when $D \to 0$ and $\epsilon \to 0$ independently; since we know that number to be at least $C_1\epsilon 8^n/(2n)$, choosing an $\epsilon_1$ small enough to have the $C\epsilon^2 8^n/ <(C_1/2)\epsilon 8^n/(2n), \epsilon<\epsilon_1$ and $D$ small enough so $DK\epsilon^4 8^n/n<(C_1/2)\epsilon 8^n/(2n)$ (here the maybe a bit imprecise $o(\epsilon^4 8^n/(n))$ is used) so $D_1 <C_1$ gives a contradiction.
Note that this argument immediately shows that the number of intervals with "many" primes (at least $D\epsilon^4 8^n/(n)$ for a small enough but fixed $D>0$ and all $\epsilon<\epsilon_1$) not only has to be at least $2$ but at least some $cK^{1/3}$, since if their number is at most $M$ then we need for example $4MC\epsilon \ge C_1$ since otherwise we can make the second term $DK\epsilon^4 8^n/n$ small enough to get a contradiction and this gives us $M \ge c_0/\epsilon \ge cK^{1/3}$
