Proof of Approximation Property of the Supremum and the Infimum in $\mathbb{R}$

Step 0 (a consequence of comparability). For $x,y\in\mathbb{R}$, if $x<y$ fails then $y\le x$. Indeed, suppose $x<y$ fails and $y\le x$ also fails. By the comparability axiom of a total order, $x\le y$. If $x=y$, then reflexivity gives $y\le x$, contrary to assumption; hence $x\ne y$, so $x<y$, again contrary to assumption. Thus $y\le x$.

Step 1 (claim 1). Assume $S$ is bounded above and $b<\sup S$, and suppose for contradiction that no $s\in S$ satisfies $b<s$. Then for every $s\in S$ the relation $b<s$ fails, so $s\le b$ by Step 0. Hence $b$ is an upper bound for $S$ in the sense of Upper Bound and Least Upper Bound. Since $\sup S$ is a least upper bound, $\sup S\le b$. Combining $b<\sup S$ with $\sup S\le b$, the mixed transitivity in claim 2 of Elementary Order Arithmetic in an Ordered Field gives $b<b$, that is $b\ne b$, which is false. This contradiction proves claim 1.

Step 2 (claim 2). Assume $S$ is bounded below and $\inf S<b$, and suppose for contradiction that no $s\in S$ satisfies $s<b$. Then for every $s\in S$ the relation $s<b$ fails, so $b\le s$ by Step 0. Hence $b$ is a lower bound for $S$ in the sense of Lower Bound and Greatest Lower Bound in a Totally Ordered Set. Since $\inf S$ is a greatest lower bound, $b\le\inf S$. Combining $\inf S<b$ with $b\le\inf S$, mixed transitivity gives $\inf S<\inf S$, which is false. This proves claim 2.

Step 3 (claim 3). Assume $S$ is bounded above. Apply the strict compatibility with addition in claim 1 of Elementary Order Arithmetic in an Ordered Field to $0<\varepsilon$ with the element $\sup S-\varepsilon$ added on the right:

$$0+(\sup S-\varepsilon)<\varepsilon+(\sup S-\varepsilon).$$

By commutativity of addition and the additive identity axiom of a field, the left-hand side equals $\sup S-\varepsilon$. By commutativity and associativity of addition together with the additive inverse and additive identity axioms,

$$\varepsilon+(\sup S+(-\varepsilon))=\sup S+(\varepsilon+(-\varepsilon))=\sup S+0=\sup S,$$

so the right-hand side equals $\sup S$. Hence $\sup S-\varepsilon<\sup S$, and claim 1 applied with $b=\sup S-\varepsilon$ produces $s\in S$ with $\sup S-\varepsilon<s$.

Step 4 (claim 4). Assume $S$ is bounded below. Apply claim 1 of Elementary Order Arithmetic in an Ordered Field to $0<\varepsilon$ with the element $\inf S$ added on the right:

$$0+\inf S<\varepsilon+\inf S.$$

By commutativity of addition and the additive identity axiom of a field, the left-hand side equals $\inf S$, and by commutativity the right-hand side equals $\inf S+\varepsilon$. Hence $\inf S<\inf S+\varepsilon$, and claim 2 applied with $b=\inf S+\varepsilon$ produces $s\in S$ with $s<\inf S+\varepsilon$.