I proved that $\left|\{n\in\mathbb{N}\colon n\leq x,\ n\text{ is squarefree}\}\right|=\frac{x}{\zeta(2)} + O(\sqrt{x})$. How does one use this result to prove that:
$\forall \theta>1/2$ $\exists$ $x_0=x_0(\theta)$ such that $\forall x\geq x_0 \,\{n\in\mathbb{N}\colon n\in[x,x+x^\theta],\ n\text{ is squarefree}\} \neq \emptyset$?
Let $F(w)$ be the number of square-frees $\le w$. There is a constant $K$ such that $$\frac{w}{\zeta(2)} -K\sqrt{w}\le F(w)\le \frac{w}{\zeta(2)} +K\sqrt{w}.\tag{1}$$ The number of square-frees in the interval $[x,x+x^\theta]$ is $\ge F(x+x^\theta)-F(x)$. By (1) this number is greater than or equal to $$\left(\frac{x+x^\theta}{\zeta(2)}-K\sqrt{x+x^\theta}\right) -\left(\frac{x}{\zeta(2)}+K\sqrt{x}\right).\tag{2}$$ We may without loss of generality assume that $\theta\le 1$. Then the number of square-frees in our interval is, by (2), greater than or equal to $$\frac{x^\theta}{\zeta(2)}-K(\sqrt{2}+1)\sqrt{x}.\tag{3}$$ Since $\lim_{x\to\infty} \frac{x^\theta}{\sqrt{x}}=\infty$, there is an $x_0$ such that (3) is greater than $0$ if $x\ge x_0$, and the result follows.