Let $\theta(x) = \sum_{p\leq x}\log p$ and $\pi(x) = |\{p\leq x:p\text{ is prime}\}|$. Using Abel's formula, one can prof the following
$$\pi(x) = \frac{\theta(x)}{\log x} + \int_{2}^{x}\frac{\theta(u)}{u(\log u)^2}du.\;\;\;\; (*)$$
Let $$\operatorname{Li}(x) = \int_{2}^{x}\frac{du}{log u} = \frac{x}{log x} - \frac{2}{log 2} + \int_{2}^{x}\frac{du}{(\log u)^2}\;\;\;\; (**)$$
If there is $c,C > 0$ such that $cx \leq \theta(x)\leq Cx$, then, using $(*)$ and $(**)$, one has
$$c\left(\operatorname{Li}(x) + \frac{2}{\log 2}\right) \leq \pi(x)\leq C\left(\operatorname{Li}(x) + \frac{2}{\log 2}\right) (***)$$
If $2c > C$. How to prove that there is $M > 0$ such that, for every $x\geq M$ there is a prime $p$ such that $x < p < 2x$? And if $2c \geq C$, there is an interval $[x,y]$ where one can guarantee the existence of a prime $p\in[x,y]$?
By using $(***)$ for estimate the diference $\pi(2x) - \pi(x)$, I think it's enough to show that there is a $M > 0$ such that
$$\int_{x}^{2x}\frac{dt}{\log t} > \frac{2}{\log 2}.$$
But I could not prove it.