I wish to show a prime in $(n,2n)$. I make a list of numbers $n+1$ through $2n-1$.
I check $n+1$ for prime factors $p_i<\sqrt{2n}$. It has a factor of $p_1$. I cross off numbers $\equiv n+1 \pmod {p_1}$.
I check $n+2$, it has $p_2$, I cross off remaining numbers $\equiv n+2 \pmod{ p_2}$. And so on, you know how sieves work.
There are only $\pi(\sqrt{2n})$ factors I have to rule out before I get a prime. Even in the naive absolute worst-case totally impossible scenario of running into a Jacobsthal $h(n)$ run of small factors and using the conservative upper bound for that, I think that still comes out to a prime popping out before $2n$, although I'll admit I haven't checked carefully, because I would hope you could rule that out anyway.
I know this is not a proof, but it seems like it could be without much work if the principle is sound; that said, the principle seems so obvious that I'm sure I'm missing something.
What am I missing?
What you are missing, is induction on $n$. Just because it works for one value of $n$, does not prove it for all other values of $n$. Proving any finite number of cases, doesn't always prove things for all $n$. That's why cover sets, and induction are so important. $n+1$ induction is a cover set for the natural numbers starting with a base case of 0 or 1. Any complete residue system for a given modulus form a trivial covering set, of the integers . All primes except the prime factors of a modulus, are covered by the equivalence classes forming the modular multiplicative group.