If $A$ and $B$ are bounded sets show that $A⊂B$ implies $\mathrm{diam}(A)≤ \mathrm{diam}(B)$ Below is what I attempt, Let $\mathrm{diam}(A) < m$ and $\mathrm{diam}(B) < n$ Let $A⊂B$ Say, $\mathrm{diam}(A)> \mathrm{diam}(B)$ $\implies m>n$ which contradicts our supposition that $A⊂B$ thus, $\mathrm{diam}(A)≤ \mathrm{diam}(B)$. Is my solution correct?

$\
ewcommand{\diam}{\text{diam}}$ $$\diam(A)=\sup\\{\|x-y\|\mid x,y\in A\\}\quad\text{and}\quad \diam(B)=\sup\\{\|x-y\|\mid x,y\in B\\}.$$ Since $A\subset B$,$$\\{\|x-y\|\mid x,y\in A\subset B\\}\subset \\{\|x-y\|\mid x,y\in B\\}.$$ The result follows. The thing you can add is the proof of $$A\subset B\implies \sup A\leq \sup B.$$