Prove that if $C \subset B$ where $B$ is a bounded subset of a metric space $(X, d)$, then $C$ is bounded and $\operatorname{diam} C \leq \operatorname{diam} B$
My Attempted Proof
Since $B$ is bounded $\exists \delta > 0$ and $x \in X$ such that $B \subseteq B_{(x, d)} (x, \delta)$
Given $C \subseteq B$ supose that $\text{diam} C > \operatorname{diam} B$. Trvially we have $\operatorname{diam} B \leq \delta$, hence $d(b_1, b_2) \le \delta$ for all $b_1, b_2 \in B$
If $\operatorname{diam} C > \operatorname{diam} B$, then there must exist $c_1, c_2 \in C$ such that $d(c_1, c_2) > d(b_1, b_2)$ for all $ b_1, b_2 \in B$, but since $C \subseteq B$ we have $c_1, c_2 \in B$ reaching a contradiction.
Therefore it follows that $\operatorname{diam} C \leq \operatorname{diam} B$
Since $B$ is bounded we have $C \subseteq B \subseteq B_{(X, d)}(x, \delta)$ and hence $C$ is also bounded. $\ \ \square$
Is my proof correct? If so how rigorous is it? Are there points where it can be improved? Any comments on my proof style and writing are greatly appreciated.
It can be simplified: the diameter proof need not be done by contradiction.
$B$ is bounded so $B \subseteq B(x, R)$ for some $X \in X, R>0$. (no need for $(X,d)$ in the subscript, as this is clear by context). Then also $C \subset B$ so the same $x$ and $R$ work to show $C$ is bounded.
Let $c_1, c_2 \in C$ then $d(c_1, c_2) \le \operatorname{diam}(B)$, as the latter number is by definition an upperbound for the set $\{d(b_1, b_2): b_1, b_2 \in B\}$and $d(c_1, c_2)$ is one of those numbers (as $C \subset B$). So $\operatorname{diam}(B)$ is an upperbound for $\{d(c_1,c_2): c_1,c_2 \in C\}$ and $\operatorname{diam}(C)$ is the by definition the smallest of these upperbounds, so $\operatorname{diam}(C) \le \operatorname{diam}(B)$