Prove that $\sup(S-T)=\sup S-\inf T$ 
Let $S$ and $T$ be nonempty sets of real numbers and define
  $$S-T=\{s-t|s\in S,t\in T\}$$
  Show that if S and T are bounded then
  $$\sup(S-T)=\sup S-\inf T\\
\inf(S-T)=\inf S-\sup T.$$

My proof:
Since $S,T\subset\mathbb{R}$ are nonempty and bounded, then, by the Completeness Axiom, we have $\alpha=\sup S,\beta=\inf T$. Let $m\in S-T$ so that $m=s-t\leq\alpha-\beta\,$ which implies that $S-T$ is bounded above so a supremum exists: denote $\gamma=\sup(S-T)\leq\alpha-\beta$. By a theorem stated in my book, pick any $\epsilon>0$, then $\exists x\in S, \exists y\in T$ such that $$(1)\,\alpha-\epsilon<x,\quad\quad(2)\,y<\beta+\epsilon\equiv-\beta-\epsilon<-y.$$
And by adding $(1)$ and $(2)$, we obtain
$$\alpha-\beta-2\epsilon\leq\alpha-\beta<x-y$$
Given that $x-y\in S-T$ then 
$$\alpha-\beta<x-y\leq\gamma.$$
Since $\gamma\leq\alpha-\beta$ and $\alpha-\beta\leq \gamma$ we have that $\gamma=\alpha-\beta$.
$$\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\square$$
Would this work?

From (2) we get $-\epsilon<\beta-y$, so $\alpha-\epsilon<\alpha+\beta-y$. The proof needs a little work, but you can salvage it. You don't need to use (1).
Also, we have $\alpha-\epsilon<x$ and $\alpha-\epsilon<\alpha+\beta-y$. This doesn't tell us $\alpha+\beta-y<x$, but we don't need that anyways.
