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?

You fix $\varepsilon > 0$ and choose $x$ and $y$ such that $\alpha - \varepsilon < x \le \alpha$ and $-\beta - \varepsilon < -y \le -\beta$. Choosing such $x$ and $y$ represent a small concession: you know that $\alpha$ and $-\beta$ may not be achievable, but you know that you can get as close as you want to these bounds, and you're happy to concede $\varepsilon$ distance from these ideals. That's why you're not going to be able to cancel these $\varepsilon$s: you can't make two of these concessions, and expect anything except that these concessions will add to each other.
Instead, observe that,
$$\alpha - \beta - 2 \varepsilon < x - y \le \alpha - \beta.$$
It follows that
$$\alpha - \beta - 2\varepsilon < \sup (A - B) \le \alpha - \beta.$$
But, there is only one number that satisfies this for any $\varepsilon > 0$. We must have
$$\sup (A - B) = \alpha - \beta.$$
