Solution: Since $S$ is a nonempty subset of $\mathbb{R} $ that is bounded below, $\inf S$ exists. Let $\alpha = \inf S$. Let us denote \[ T = \{ -s : s \in S \}. \] We claim that $-\alpha $ is the least upper bound (supremum) of the set $T$. First observe that for any $t \in T$, there exists $s \in S$ such that $t = -s $. Since $\alpha$ is a lower bound of $S$, we have \begin{align*} s \geq \alpha \implies -s \leq -\alpha \implies t \leq -\alpha . \end{align*} As $t$ was arbitrarily chosen, we have shown that $-\alpha $ is an upper bound of $T$.

We now will show that $-\alpha $ is the least upper bound of $T$. Let $\beta $ be an upper bound of $T$. Then for any $t = -s \in T$, we have \begin{align*} t \leq \beta \implies -s \leq \beta \implies s \geq -\beta . \end{align*} As $\alpha $ is the greatest lower bound (infimum) of $S$, we have \[ -\beta \leq \alpha \implies \beta \geq -\alpha . \] Thus, $\alpha $ is the least upper bound of $T$ and hence \[ \inf S = -\sup \{ -s: s \in S \}. \]