Suppose $ X \subset \mathbb{R},\ $ with $\ \vert X\vert \geq 3, $ has the property that:
$$ \forall\ x,y \in X\ \text{ with} \ x<y,\ \exists\ z\in X\ \text{ such that } x<z<y.\qquad (\text{Property } 1)$$
Proposition: $\ X\ $ is somewhere dense, that is: $\exists\ a,b, $ with $ a<b\ $ such that $ [a,b] \subset \overline{X}. $
Note that Property $1$ does not say that $X$ is connected.
I was thinking something like, the Cantor set with endpoints removed could be a counter-example, but I'm not sure. And I also don't see how to proceed in a proof, for example a proof-by-contradiction. If we started assuming (by contradiction) that $X$ is nowhere dense, then this does not mean that every point of $X$ is an isolated point. So I am a bit stuck, although it is possible I am missing something simple.
Consider the set of numbers which does not have $1$, in it’s ternary expansion and also has a unique ternary expansion, it clearly has the between-ness property but it is nowhere dense. In fact this is almost exactly the Cantor set.