Analysis (Calculus) proof regarding inequalities, sup/inf

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Homework Statement

1. Let $S$ be a nonempty set of real numbers that is bounded from below. Define $T=\{-s: s \in S\}$. Prove that $T$ is nonempty and bounded from above. Then prove that $\sup(T)=-\inf(S)$.\\

The Attempt at a Solution

$S$ is nonempty
so there is some $s \in S$ and also $s \in R$. \\
All the elements in R have an additive inverse so with $s \in R$ there is $-s \in R$ and $-s \in T$. Thus T is nonempty. \\
S is bounded from below so there is an $inf(S)$ so for $\forall s \in S$, $s \geq inf(S)$
$\Rightarrow$ $-s \leq -inf(S) \Rightarrow$ \\ $\forall t \in T$ $t \leq -inf(S)$. So $-inf(S)$ is an upper bound for T.
\\
How do you prove it's the lowest upper bound?

Say that $sup T < -inf S$ $\Rightarrow$ $-sup T > inf S$ $\Rightarrow$ there's an element smaller than -sup T also in S, call it n.
$n \in S \Rightarrow -n \in T$
$n < -sup T \Rightarrow -n > sup T$
$-n \in T$ and $-n >sup T$
Contradiction, therefore $sup T=-inf S$

let c be a lower bound, then $$c \leq x$$ for $$\forall x \epsilon S$$. From here, you have to use algebra to get -c in the inequality somehow*. Use this to show that -c is an upper bound for S.
Then, use Axiom of Completeness, there exists a b=supS such that for $$b \leq c$$ for $$\forall x \epsilon S$$ (same as from *) to continue the proof.