Let $A$ and $B$ be nonempty bounded subsets of $\mathbb{R}$. Define $|A-B|=\{|x-y|: x\in A, y\in B\}$. I wish to prove the following claim.
Claim: $$|\sup A-\sup B|\leq \sup |A-B|$$
Is the following a correct proof?
Proof Let $a=\sup A$ and $b=\sup B$. Let $\epsilon >0$ and by definition of supremum pick $x\in A$ such that $a-\frac\epsilon2< x\leq a$ and similarly pick $y\in B$ so that $b-\frac\epsilon2 <y\leq b$. Then it follows that $|(x-y)-(a-b)|\leq \frac\epsilon2 <\epsilon$ and by the arbitrariness of $\epsilon>0$ it follows that $(x-y)=a-b$ and so $|x-y|=|a-b|$. But we have $x\in A$ and $y\in B$ which implies $|x-y|\in |A-B|$. So since $|a-b|=|x-y|$ we have $|a-b|\in |A-B|$ hence by definition of supremum we obtain $|a-b|\leq \sup |A-B|$. The result follows since $a=\sup A$ and $b=\sup B$.
Am I making a subtle or obvious mistake anywhere or is that all there is to it?
Alternatively: What are some other ways (read: hints) to prove this?
There is one subtle mistake. You say that by the arbitrariness of $\epsilon$ you can assume that $x - y = a - b$. Notice that this is not quite correct since both $x$ and $y$ depend on $\epsilon$, so when sending $\epsilon$ to $0$ you might be changing these two points and end up with something that in the limit is no longer in $A$ or $B$.
Luckily there is an easy fix: using the definition of supremum (exactly as you did!) we show that for every $\epsilon > 0$ there are $x \in A$ and $y \in B$ such that, by the triangle inequality, $$|a - b| \le |a - x| + |x - y| + |y - b| \le |x - y| + \epsilon \le \sup|A-B| + \epsilon.$$ No you can finally let $\epsilon \to 0^+$.