I don't know if I am right. So I came here. What I did was use the theta function, defined as the sum $\theta(x)=\sum_{p\leq x} \log{p}$. So we have that there is always a prime between x and 2x, for x sufficiently large. Taking this function on both sides yields $\theta(x)$<$\theta(2x)$. What I did was this. I wrote $\theta(x)$ in terms of the previous sum. So in particular, $\theta(x)=\sum_{p\leq 2x} \log{p}$=($\sum_{p\leq x} \log{p}$)+($\sum_{p<q\leq 2x} \log(q)$). So in particular, there is some $\log q$ > $\log p$, where the $\log(p)$ is "contained" in $\theta(x)$, and $\log(q)$ is "contained" in the larger sum, $\theta(2x)$. So we have that there exists a $\log q$ in between $\theta(x)$ and $\theta(2x)$. (Or more exactly, the logarithms in the sum). So we have there is a prime between x and 2x.
I know its worded badly, but I suppose would could say it is a motivation for a more formal version, for which I am still working on. But I hope it makes sense up to some degree.