Proof of An Open Interval is an Interval All of Whose Points Are Interior

We use the elementary order arithmetic of the ordered field $\mathbb{R}$; claim numbers below refer to that lemma.

Step 0 (midpoints). If $s,t\in\mathbb{R}$ satisfy $s<t$, then the element $m=(s+t)\cdot 2^{-1}$ satisfies $s<m$ and $m<t$. Indeed, $2^{-1}$ exists and $0<2^{-1}$ by claim 8. From $s<t$, claim 1 (adding $s$) gives $s+s<s+t$, and claim 10 (multiplying by $2^{-1}$) gives $(s+s)\cdot 2^{-1}<(s+t)\cdot 2^{-1}=m$; since $s+s=s\cdot(1+1)=s\cdot 2$ and $(s\cdot 2)\cdot 2^{-1}=s$, this reads $s<m$. Symmetrically, claim 1 (adding $t$) gives $s+t<t+t$, and claim 10 gives $m<(t+t)\cdot 2^{-1}=t$.

Step 1 ($(p,q)$ is an interval). Let $x,z\in(p,q)$ and let $y\in\mathbb{R}$ satisfy $x\le y$ and $y\le z$. From $p<x$ and $x\le y$, claim 2 gives $p<y$; from $y\le z$ and $z<q$, claim 2 gives $y<q$. Hence $y\in(p,q)$, which is the defining condition of an interval.

Step 2 (every point is interior). Let $x\in(p,q)$, so $p<x$ and $x<q$. Put $u=(p+x)\cdot 2^{-1}$ and $v=(x+q)\cdot 2^{-1}$. By Step 0 applied to $p<x$ we get $p<u$ and $u<x$; by Step 0 applied to $x<q$ we get $x<v$ and $v<q$.

Then $u\in(p,q)$: we have $p<u$, and $u<x$ with $x<q$ gives $u<q$ by claim 2. Likewise $v\in(p,q)$: we have $v<q$, and $p<x$ with $x<v$ gives $p<v$ by claim 2.

So $u,v\in(p,q)$ with $u<x$ and $x<v$, which is exactly the defining condition for $x$ to be an interior point of $(p,q)$.