This is what I have so far (I don't know if this is the best approach):
Since $x\geq 1$, we know that for all $q\in\mathbb{Q}$ with $q<p$, we have $x^q\leq x^p$.
Taking supremums on $q$, we get $\sup\{x^q\mid q\in\mathbb{Q},q<p\}\leq x^p$.
By contradiction, suppose that there exists $\varepsilon>0$ such that $x^q+\varepsilon\leq x^p$ for all $q\in\mathbb{Q}$ with $q<p$. Let's write $p:=\dfrac{m_1}{n}$ and $q:=\dfrac{m_2}{n}$. Then, $(x^q+\varepsilon)^n\leq x^{m_1}$. Therefore, $\left(1+\dfrac{\varepsilon}{x^q}\right)^n\leq\dfrac{x^{m_1}}{x^{m_2}}$.
Using Bernouilli's inequality, we've got $1+n\cdot\dfrac{\varepsilon}{x^q}\leq\dfrac{x^{m_1}}{x^{m_2}}$.
I can't reach any contradiction from here.
Has someone any suggestion?
Thanks in advance.