Finishing up baby Rudin ch. 2, I was doing problems and realized that I had not seen a proof that given $p \in \mathbb{R}^n$, $q \in \mathbb{R}^n$, $d(p,q) > 0$, there exists another point $z \in \mathbb{R}^n$ such that $0 < d(p,z) < d(p,q)$. In fact, the first proof relating to $\mathbb{R}^n$ in Rudin, that all open (or closed) balls are convex, states that if $|y-x| < r$, and $|z-x| < r$, and $0<\lambda<1$, then the result of $\lambda x + (1-\lambda)y$ is in that open ball as well. But this only states that such a point would be, the existence of two such points in an open ball is not assumed (or even necessary).
I've spent some time looking for evidence of this proposition in chapter 2, but I haven't found anything yet; the only contenders seem to be the proofs of compactness, but I think they hold without it (although they do depend on the LUB property of the reals). At first I thought I was being too pedantic, or that this should be trivial to prove, but I tried to sketch some ideas (other than directly showing that you can solve for such a point $z$ between $p$ and $q$ by the distance metric formula) but I didn't come up with anything right away. I don't think it's a hard problem, but now I'm getting curious if maybe, in Rudin, this was avoided on purpose as an unnecessary complication, or was tacitly assumed?
The completeness of the reals I hope we can assume, and it comes up in any first real analysis course. Then you can argue inductively to extend the result to $\mathbb{R} ^n$, I reckon.
Then what you can do is say, assume that $\mathbb{R} ^n$ is complete. Then for $\mathbb{R} ^{n+1}$ we have that this is a direct product of two complete spaces. So for any a, b we have that we can take $a+ \lambda *(b-a)$. As the first n co-ordinates are in $\mathbb{R} ^n$, and the last is in $\mathbb{R}$ we have that it exists.
This question is also, to some extent, about the reals being hausdorff, and as you have a metric on this space, they are hausdorff (all metric spaces are.) Though I do feel you're looking for a more tangible solution than "it is".