Legendre's conjecture states that for all $n$, there is a prime number between $n^2$ and $(n+1)^2$.
It has been proven that $\lim_{n\to\infty}\pi(n) = \frac{n}{\ln(n)}$.
It can be proven that $\frac{(n+1)^2}{\ln((n+1)^2)} – \frac{n^2}{\ln(n^2)} > 1$ for all $n$.
At some large $n$, the difference between $\pi((n+1)^2) - \pi(n^2))$ and $\frac{(n+1)^2}{\ln((n+1)^2)} – \frac{n^2}{\ln(n^2)}$ would be small enough to confirm Legendre's conjecture.
No, because the error terms can hide a lot. You would really need to show that $$ (1-\varepsilon)\frac{(n+1)^2}{\ln((n+1)^2)} – (1+\varepsilon)\frac{n^2}{\ln(n^2)} > 1 $$ when $n$ is large enough in terms of $\varepsilon$, but this is not true.
Here's a more concrete example: let $f(x)$ denote the number of integers up to $x$ whose last digit is either $1$, $2$, or $3$. It's easy to prove that $f(x)$ is asymptotic to $3x/10$ as $x\to\infty$. Note that $3(n+4)/10 - 3n/10 > 1$ for all $n$ . Does that show that there is always a number between $n$ and $n+4$ whose last digit equals $0$, $1$, or $2$?