Ok sorry, I was a bit high.

So what's your construction? I think that neither Dirichlet, nor CRT can give such construction.

Sketch: Just work in mod 5. we consider an integer i such that p(i)=p(i+4) (mod 5) or p(i)=p(i+3) (mod 5). Let a(j)=p(i+j)-p(i). Case 1: p(i+4)=p(i) (mod 5): Then a(1)+a(2)+a(3)+a(4)=0 (mod 5), it is easy to see that {a(1),a(2),a(3),a(4)}={6,-12,6,12}, so there are 2 equal numbers. Case 2: p(i+3)=p(i) (mod5): Then a(1)+a(2)+a(3)=0 (mod 5), it is easy to see that a(1)=12, a(2)=6, a(3)=12 or a(1)=-12, a(2)=-6, a(3)=-12, otherwise there are 2 equal numbers. WLOG a(1)=12, a(2)=6, a(3)=12 Then p(i+1)=p(i)+12, p(i+2)=p(i)+18, p(i+3)=p(i)+30 Consider p(i+1), p(i+2), p(i+3), p(i+4), p(i+5). If p(i+1)=p(i+4) (mod 5) then there are 2 equal numbers. Then p(i+2)=p(i+5) (mod 5) and we have p(i+4)=p(i)+36, p(i+5)=p(i)+48 Similarly, we have p(i+6)=p(i)+54, p(i+7)=p(i)+66. Consider p(i), p(i)+12, p(i)+18, p(i)+48, p(i)+36 and p(i)+54, at least one of these number is divisible by 5, contradiction because they are primes. Done.

We can actually solve the problem with CRT. Take some prime $p \geq 11$ from the given. Now there exists $1 \leq k \leq 35$, such that $p+6k$ and $p+6(k+1)$ are divisible by $5$ and $7$ (we can actually take some other pair primes since $2021$ is unnecessarily large, but $5$ and $7$ are the minimal possible). Now we have two adjacent numbers with difference at least $18$ (namely $p+6(k-1)$ and $p+6(k+2)$), contradiction.