Here is the problem in its entirety:
Prove that every non-empty $S\subseteq \mathbb{R}$ that has a lower bound has a greatest lower bound. For every set $S$ it will be helpful to think about the associated set $A=\{-x:x\in S\}$.
Using the definition where if $S\subseteq \mathbb{R}$ and $l\in \mathbb{R}$ we say that $l$ is a $\it lower$ $\it bound$ $\it for$ $\it S$ if $\forall x\in S$, $l \leq x$. Also, $\gamma \in \mathbb{R}$ is a $\it greatest$ $\it lower$ $\it bound$ $\it for$ $\it S$ if $\gamma$ is a lower bound for $S$, and moreover $\forall l\in \mathbb{R}$, $l$ is a lower bound for $S$ $\implies$ $l \leq \gamma$.
a.) Prove that for all $S\subseteq \mathbb{R}$, and all $l\in \mathbb{R}$, $l$ is a lower bound for S $\implies$ $-l$ is an upper bound for $A$.
b.) Prove that for all $S\subseteq \mathbb{R}$, and all $u\in \mathbb{R}$ , $u$ is an upper bound for $A$ $\implies$ $-u$ is a lower bound for $S$.
c.) Prove that for all non-empty $S\subseteq \mathbb{R}$ having a lower bound, $S$ has a greatest lower bound, and glb($S$) $= -$lub($A$).
I have proved least upper bounds before and I know that within this I need to show that $A$ has an upper bound, so that lub($A$) is defined, and that I then need to prove that -lub($A$) is a lower bound and that its at least as big as any other lower bound, but I'm having trouble with how to get started and write down these individual parts to a,b,c.
a.) If $l \in \mathbb R$ is a lower bound for $S\subseteq \mathbb{R}$, then $\forall x \in S, l \le x,$ i.e., $\forall x \in A, l \le -x,$ i.e., $-l \ge x,$ so $-l$ is an upper bound for $A$.
b.) If $u \in \mathbb R$ is an upper bound for $A$, then $\forall x \in A, x \le u,$ i.e., $\forall x \in S, -x \le u$, i.e., $\forall x \in S, x \ge -u,$ so $-u$ is a lower bound for $S$.
c.) Say $S$ has a lower bound; call it $l.$ Then by a.), $-l$ is an upper bound for $A$. Therefore, by completeness, assuming $S$ is not empty and therefore $A$ is not empty, $A$ has a lub; say $u=$ lub($A$). Since $u$ is an upper bound for $A$, it follows from b.) that $-u$ is a lower bound for $S$. We now wish to show that $-u$ is glb($S$). Let $z$ be any lower bound for $S$. Therefore, by a.) $-z$ is an upper bound for $A$. Since $u$ is the lub of $A$, $u\le-z$. Therefore $-u\ge z;$ i.e., $-u=$ glb$(S).$