My approach starts out similar to @CliveNewstead's but then I let $-\gamma = \sup(-A)$ and used the fact that the infimum is unique.
Since $A$ is nonempty and bounded below, $A = \{a:a\in A\}$ and $\inf(A) = \alpha$. Now, $-A = \{-a:a\in A\}$ is also nonempty. Since $\alpha$ is the infimum of $A$, $\alpha\leq a$ for all $a\in A$. By multiplying by $-1$, we get the following inequality
$$
-\alpha\geq -a.
$$
That is, $-\alpha$ is an upper bound of $-A$. Suppose $-\gamma = \sup(-A)$ and $\varepsilon > 0$. Then $-\gamma + \varepsilon\not\in -A$
$$
-\alpha\geq -\gamma + \varepsilon\geq -\gamma\geq -a
$$
Again, by multiplying by negative one, we have
$$
\alpha\leq \gamma - \varepsilon\leq\gamma\leq a
$$
but $\gamma - \varepsilon\notin A$ so $\gamma$ is a lower bound of $A$ which would contradict the fact that $\alpha$ is the greatest lower bound of $A$. In order for $\gamma$ to be the lower bound, $\gamma = \alpha$ since the infimum is unique. So $-\alpha$ is the supremum of $-A$. Therefore, $\alpha = \inf(A) = -\sup(-A) = -(-\alpha) = \alpha$.