I tried a proof using mainly the Archimedean property of $\Bbb R$:
I am to show that for any $x\in \Bbb R$, every neighborhood of $x$ contains a rational number. I shall first prove it for non-negative real numbers. Suppose $x$ is a non-negative real number and let $\epsilon >0$ be arbitrary. By the Archimedean property, there is an $n\in \Bbb N$ such that $n\epsilon >1$ i.e. $\frac 1n<\epsilon$. Again, using the Archimedean property, we can find an $m\in \Bbb N$ such that $m\frac 1n >x$.
Let $S=${$m\in \Bbb N : \frac mn>x$}. Since $S$ is a nonempty subset of $\Bbb N$, it has a least element, say, $m'$. It means one of the following:
A) $m'-1$ is a natural number with the property that $\frac {m'-1}n≤x$; or
B) $m'-1=0$ whence $m'=1$.
A) If possible assume that neither $\frac {m'-1}n$ nor $\frac {m'}n$ belong in $(x-\epsilon,x+\epsilon)$. Which means :
$\frac{m'-1}n<x-\epsilon<x+\epsilon<\frac{m'}n \implies (x+\epsilon) - (x-\epsilon)<\frac{m'}n - \frac{m'-1}n \implies 2\epsilon<\frac 1n<\epsilon$ which is a contradiction. Thus one of them must be in $(x-\epsilon,x+\epsilon)$.
B) $m'=1$ means that $\frac 1n>x$. But since $x$ is non-negative we also have $\frac 1n<\epsilon<x+\epsilon$. Hence, $\frac 1n\in (x-\epsilon,x+\epsilon)$.
Thus every neighborhood $(x-\epsilon,x+\epsilon)$ of $x$ contains a rational number $q$. But then, the neighborhood $(-x-\epsilon,-x+\epsilon)$ of $-x$ contains the rational number $-q$. This completes my proof.
Is the proof okay?