Proof of Existence of the Infimum of a Nonempty Subset of $\mathbb{R}$ Bounded Below

Let $L$ be the set of all lower bounds of $S$ in the sense of Lower Bound and Greatest Lower Bound in a Totally Ordered Set, that is,

$$L=\{\ell\in\mathbb{R}:\ \ell\le s\ \text{for every}\ s\in S\}.$$

Step 1: $L$ is nonempty. By hypothesis $S$ is bounded below, so some lower bound of $S$ exists, and that element lies in $L$.

Step 2: every element of $S$ is an upper bound for $L$, and $L$ is bounded above. Let $s\in S$ and $\ell\in L$. By the definition of $L$ we have $\ell\le s$. Since $\ell\in L$ was arbitrary, $s$ is an upper bound for $L$ in the sense of Upper Bound and Least Upper Bound. As $S$ is nonempty there is at least one such $s$, so $L$ is bounded above.

Step 3: $L$ has a least upper bound. By Steps 1 and 2, $L$ is a nonempty subset of $\mathbb{R}$ that is bounded above, so by the least upper bound property in The Real Numbers it has a least upper bound in $\mathbb{R}$; call it $m$.

Step 4: $m$ is a lower bound for $S$. Let $s\in S$. By Step 2, $s$ is an upper bound for $L$. Since $m$ is a least upper bound for $L$, it satisfies $m\le b$ for every upper bound $b$ of $L$; taking $b=s$ gives $m\le s$. As $s\in S$ was arbitrary, $m$ is a lower bound for $S$.

Step 5: $m$ is a greatest lower bound for $S$. Let $\ell$ be any lower bound of $S$. Then $\ell\in L$ by the definition of $L$, and $m$ is an upper bound for $L$, so $\ell\le m$.

By Steps 4 and 5 and Lower Bound and Greatest Lower Bound in a Totally Ordered Set, $m$ is a greatest lower bound of $S$ in $\mathbb{R}$. Since $\le$ is a total order on $\mathbb{R}$, Uniqueness of the Supremum and of the Infimum shows $m$ is the only one, so writing $\inf S=m$ is unambiguous.