Proof your answer(asumption) $A$ is a bounded subset of $(0, \infty)$. $B$ is a subset of  $\mathbb{R}$ defined by $B=\{{\sqrt{a_1}-a_2} \text{ such that }a_1,a_2 \in A\}$. Compute $\sup(B)$ in terms of $\sup(A)$ and $\inf(A)$. 
 A: Hint: $\sup B=\sup\sqrt{A}-\inf A$. Intuitively, this is because $\sqrt{a}$ is an increasing function while $-a$ is a decreasing function. Can you prove the claim formally?

Here's the full answer, if you are still having trouble. It is more verbose than is ideal, but hopefully it leaves no stone left unturned.
Since $A$ is bounded and nonempty, we can find sequences $(\overline{a}_{n})_{n}$
and $(\underline{a}_{n})_{n}$such that $\overline{a}_{n}\rightarrow\sup A$
and $\underline{a}_{n}\rightarrow\inf A$. This allows us to construct
the sequence $(b_{n})_{n}$ given by $b_{n}=\sqrt{\overline{a}}_{n}-\underline{a}_{n}$.
Since limits and continuous functions commute,
$$
\lim_{n\rightarrow\infty}b_{n}=\lim_{n\rightarrow\infty}\sqrt{\overline{a}_{n}}-\underline{a}_{n}=\sqrt{\lim_{n\rightarrow\infty}\overline{a}_{n}}-\lim_{n\rightarrow\infty}\underline{a}_{n}=\sqrt{\sup A}-\inf A.
$$
It follows that $\sup B\geq\sqrt{\sup A}-\inf A$.
Now, to arrive at a contradiction, suppose $\sup B>\sqrt{\sup A}-\inf A$.
It follows that there exists some $b$ in $B$ and corresponding such
that $b>\sqrt{\sup A}-\inf A$. By the definition of $B$, this implies
the existence of $a_{1}$ and $a_{2}$ in $A$ such that $\sqrt{a_{1}}-a_{2}>\sqrt{\sup A}-\inf A$.

Once you get used to this argument, you can skip the details and simply
write 
$$
\sup B=\sup_{a_{1},a_{2}\in A}\sqrt{a_{1}}-a_{2}=\sup_{a_{1}\in A}\sqrt{a_{1}}+\sup_{a_{2}\in A}(-a_{2})=\sup_{a_{1}\in A}\sqrt{a_{1}}-\inf_{a_{2}\in A}a_{2}=\sup A-\inf A
$$
