For the case $0 < a < b$ which is what I'm interested in, there is a proof that there exists a rational number on the open interval which I've seen many times but I don't really understand it.
I accept it's valid as I've seen it from reliable sources but I don't "get it". A brief expository statement to help me see why it works, as it's not obvious to me, would be appreciated.
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A possible intuition. Consider the decimal sequences of $$ a = \ldots a_{-2} a_{-1}. a_1 a_2 \ldots\\ b = \ldots b_{-2} b_{-1}. b_1 b_2 \ldots\\ $$
Find the first place where they differ, say $b_k$. Then the number $$ k = \ldots b_{-2} b_{-1}. b_1 b_2 \ldots b_k $$ is rational and is in $(a,b)$.
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Let me start by saying that every real number has a sequence of rationals approaching it. For example, $\pi$ is approached by the following sequence: $$3,3.1,3.14,3.141,\dots$$ In general, any real number can be approached by rationals by taking the first $n$ digits.
Let $M=\dfrac{a+b}2$, and let $M_n$ be a sequence of rationals approaching $M$. Since they approach it, they eventually get within a distance of $\epsilon$ of it for any $\epsilon$. Simply let $\epsilon$ be $b-M=M-a$.
The idea is that you choose a numerator $n$ that is large enough that $1/n < b-a$. In that case, the distance between consecutive fractions like $\frac in, \frac{i+1}n$ is smaller than the width of the interval $(a, b)$. That means that at least one of those fractions must be inside the interval.