This is quite trivial if viewed correctly. In a nutshell it boils down to the observation that we can employ rational-based rulers, i.e. we can change our unit of measurement on the number line from $1$ to any rational $\rm\:s\:$. More precisely, suppose we wish to find a rational in the real interval $\rm\ (a,b),\ a,b\in \mathbb R\:$. Simply choose some rational $\rm\:r < a\ $ and a positive rational step size $\rm\ s < b-a\:.\ $ Now, starting from $\rm\:r\:,\ $ keep taking steps of size $\rm\:s\:.\ $ By the Archimedean Property (AP) eventually you'll surpass $\rm\:b\:$, and taking a step back from the first such point necessarily lands you at a rational point in $\rm\ (a,b)\:,\ $ since the step size is smaller than the interval. Since the proof uses only the Archimedean property it works for any Archimedean field. Essentially it's simply employing the (Euclidean) division algorithm for rationals. Equivalently, instead of choosing the unit $1$ for our number line we instead choose the unit to be the rational number $\rm\:s\:$ that is our step size. In fact these ideas go all the way back to Euclid, who used it to (effectively) compute the gcd of rationals (greatest common measure of line segments).
It proves illuminating to examine the proof in JM's answer from this geometric viewpoint: by AP we can scale the interval $\rm\ I = (a,b)\ $ by an integer factor $\rm\:n\:$ to obtain $\rm\:n\:I = (na,nb)$ of length $> 1\:.\ $ Thus by AP there is an integer $\rm\:k\in n\:I\:$ hence a rational $\rm\:k/n\in I\:.\ $ QED $\ $ Notice how much more intuitive the proof is when presented in this geometric form. The key to its success is that the problem is invariant under rational scaling symmetries. Thus we can scale it to reduce it to what amounts to a trivial floor or ceiling computation (these functions exist by AP and induction).