I've seen this problem mentioned on Math.SE before, but not proved thoroughly. With that said I wanted to attempt it for myself and have my work verified.
Let $(M,d)$ be a metric space and $A \subset M$. If $\operatorname{diam}(A) < r$, show that $A \subset B_{r}(a)$ for some $a \in A$.
The proof for this seems pretty short:
Let $a_{o}, b \in A$ and $r > 0$. Since $\operatorname{diam}(A) := \sup\{d(a,b): a, b \in A\} < r$, then $d(a_o,b) < \operatorname{diam}(A) < r$. So since $d(a_{o},b) <r$ by definition of an open ball of radius $r$, $b \in B_{r}(a_{o})$ for some $a_{o} \in A$. Therefore $A \subset B_{r}(a_{o})$.
Any criticism is welcome.
Best Answer
It is almost correct. The only (minor) problem is when you state that $d(a_0,b)<\operatorname{diam}(A)$. This is not true; all you can say is that $d(a_0,b)\leqslant\operatorname{diam}(A)$. It it still follow from this that $d(a_0,b)<r$.