Proof
(related to Proposition: Rational Numbers are Dense in Real Numbers)
 By hypothesis, $x\in\mathbb R$ is a real number and $\epsilon > 0$ any (arbitrarily small) positive real number.
 We are going to find a rational number $q\in\mathbb Q$ with $q\in(x\epsilon, x+\epsilon)$ using a constructed convergent rational sequence.
 First, we choose two rational numbers $\frac {a_0}1,\frac{b_0}1\in\mathbb Q$ with some integers $a_0 < x < b_0.$ This is possible because of the Archimedean axiom.
 Note that $x\in [a_0,b_0],$ where $[a_0,b_0]$ is a closed interval.
 Now, we construct for $n=1,2,\ldots$ new closed intervals by the following rule:
* if $\frac{a_n+b_n}2 > x$, we set $a_{n+1}:=a_n$ and $b_{n+1}=\frac{a_n+b_n}2,$
* otherwise, we set $a_{n+1}:=\frac{a_n+b_n}2,$ and $b_{n+1}:=b_n.$
 By construction, the sequences $(a_n)_{n\in\mathbb N}$ and $(b_n)_{n\in\mathbb N}$ are rational sequences with $x\in [a_{n},b_{n}]$ for all $n:=1,2,\ldots.$
 Since the length of the $n$th interval is $(b_0a_0)/2^n,$ the sequences are also real convergent sequences with the limit $$x=\lim_{n\to\infty} a_n=\lim_{n\to\infty} b_n.$$
 Although the limit $x$ does not have to be a rational number^{1} itself, all sequence members $a_n,b_n$ are, by construction, rational numbers for all $n\in\mathbb N.$
 From the definition of convergence, it follows that, taking $(a_n)_{n\in\mathbb N}$ as an example, there is an index $N\in\mathbb N$ such that $a_nx < \epsilon$ for all $n > N.$
 By setting $q:=a_N\in (x\epsilon, x+\epsilon),$ we have found a rational number that lies arbitrarily dense with the real number $x.$
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References
Bibliography
 Modler, F.; Kreh, M.: "Tutorium Analysis 1 und Lineare Algebra 1", Springer Spektrum, 2018, 4th Edition
Footnotes