I find this result without a proof:
The prime number theorem implies that for any $\epsilon > 0$, there exists an integer $N$ such that for all $x > N$, there exists a prime between $x$ and $x(1 + \epsilon)$.
I am asking about a simple proof for this result.
By prime number theorem, there are approximately $\frac{N}{\ln(N)}$ primes less than $N$, for sufficiently large $N$. Therefore, between $x$ and $x(1+\epsilon)$ there are approximately $$\frac{x(1+\epsilon)}{\ln(x(1+\epsilon))}-\frac{x}{\ln(x)}$$ primes. This expression is equivalent to $$\frac{x(1+\epsilon)\ln(x)-x\ln(x(1+\epsilon))}{\ln(x(1+\epsilon))\ln(x)}$$ which, by L'Hopital's rule, approaches $\infty$ as $x\to\infty$.
In particular, this means there is some $N$ such that for all $x>N$ we have at least one prime between $x$ and $x(1+\epsilon)$.