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?
 A: $\newcommand{\diam}{\text{diam}}$
$$\diam(A)=\sup\{\|x-y\|\mid x,y\in A\}\quad\text{and}\quad \diam(B)=\sup\{\|x-y\|\mid x,y\in B\}.$$
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 $$A\subset B\implies \sup A\leq \sup B.$$
A: $\newcommand{\diam}{\operatorname{diam}}$$A\subset B$ means every member of $A$ is a member of $B$.  That is a definition.
$\diam(A)$ is usually defined as $\sup\{d(x,y) : x,y\in A\}$.  That is a definition.
Recall also the definition of "upper bound" and "smallest upper bound".
If $A\subset B$ then every distance between two points in $A$ is a distance between two points in $B$, since if the two points are in $A$, then they are in $B$, by the first definition above.  Therefore
$$
\{d(x,y):x,y\in A\} \subset \{d(x,y): x,y\in B\}.
$$
The number $\diam(B)$ is the smallest upper bound of the latter of those two sets.  That implies $\diam(B)$ is an upper bound of $B$.  That means $\diam(B) \ge\text{every distance between two points of }B$.  But every distance between two points of $A$ is a distance between two points of $B$, since all points in $B$ are in $A$.  Hence $\diam(B)$ is $\ge\text{all distances between points of } A$.  Thus $\diam(B)$ is an upper bound of the set of all distances between points of $A$.  Hence $\diam(B)\ge \text{the smallest upper bound of all distances between points of } A$.  The latter smallest upper bound is $\diam(A)$.  So $\diam(B)\ge\diam(A)$.
A: Sorry, but your attempt is not good at proving what you're required to. You're basically proving that, if $m>n$ then it is false that $m\le n$.

Let's see what $\def\diam{\operatorname{diam}}\diam(A)$ is. Consider the set $A^*$ of all numbers you obtain by doing $d(x,y)$ where $x,y\in A$ ($d$ is the distance function on the space you're working in).
Then $\diam(A)=\sup A^*$, by definition.
Now, if $A\subset B$, that is, every point in $A$ is also in $B$, we get immediately that $A^*\subset B^*$, so our real task is showing that, if $P$ and $Q$ are sets of real numbers such that $P\subset Q$, then $\sup P\le\sup Q$.
This will follow from the observation that any upper bound for $Q$ is also an upper bound for $P$. Therefore the least upper bound for $Q$ is an upper bound for $P$, hence it is greater than or equal to the least upper bound for $P$. In other words, $\sup P\le \sup Q$.
