[proofplan] We use the definitions directly. If $A$ is bounded, then all points of $A$ lie within a fixed finite distance of one center, so the metric triangle inequality gives a uniform bound on every pairwise distance in $A$. Conversely, if the pairwise distances in $A$ have finite supremum, then choosing one point $a_0\in A$ gives a finite-radius ball centered at $a_0$ containing all of $A$. [/proofplan] [step:Bound every pairwise distance from a common finite-radius center] Assume that $A$ is bounded. By definition, there exist a point $x_0\in X$ and a real number $R\ge 0$ such that \begin{align*} d(a,x_0)\le R \end{align*} for every $a\in A$. Define the set of pairwise distances \begin{align*} S:=\{d(a,b):a,b\in A\}\subset [0,\infty). \end{align*} Let $a,b\in A$. Since $d$ is a metric, the triangle inequality gives \begin{align*} d(a,b)\le d(a,x_0)+d(x_0,b). \end{align*} Because $d$ is symmetric and $a,b\in A$, we have \begin{align*} d(a,x_0)+d(x_0,b)=d(a,x_0)+d(b,x_0)\le R+R=2R. \end{align*} Thus every element of $S$ is at most $2R$, so $S$ is bounded above. Since \begin{align*} \operatorname{diam}(A):=\sup S, \end{align*} we obtain \begin{align*} \operatorname{diam}(A)\le 2R<\infty. \end{align*} [/step] [step:Use finite diameter and nonemptiness to choose a center inside $A$] Assume that \begin{align*} \operatorname{diam}(A)<\infty. \end{align*} Because $A$ is nonempty, choose a point $a_0\in A$. Define \begin{align*} D:=\operatorname{diam}(A). \end{align*} Then $D<\infty$. For every $a\in A$, the distance $d(a,a_0)$ belongs to the set \begin{align*} \{d(u,v):u,v\in A\}, \end{align*} because $a\in A$ and $a_0\in A$. Hence, by the defining property of the supremum, \begin{align*} d(a,a_0)\le D. \end{align*} Therefore every point of $A$ lies within finite distance $D$ of the point $a_0$, so $A$ is bounded. [guided] Assume that \begin{align*} \operatorname{diam}(A)<\infty. \end{align*} The definition of boundedness asks for one center and one finite radius controlling all points of $A$. The finite diameter controls distances between pairs of points of $A$, so we need one fixed point of $A$ to serve as the center. This is exactly where the nonempty hypothesis is used: since $A\ne\varnothing$, choose a point $a_0\in A$. Define \begin{align*} D:=\operatorname{diam}(A). \end{align*} By assumption, $D<\infty$. The diameter is the supremum of all pairwise distances in $A$, namely \begin{align*} D=\sup\{d(u,v):u,v\in A\}. \end{align*} Now let $a\in A$ be arbitrary. Since both $a$ and $a_0$ belong to $A$, the number $d(a,a_0)$ is one of the pairwise distances used in the supremum defining $D$. Therefore the defining property of a supremum gives \begin{align*} d(a,a_0)\le D. \end{align*} Because this holds for every $a\in A$ and $D$ is finite, all points of $A$ lie in the finite-radius closed ball centered at $a_0$. Hence $A$ is bounded. [/guided] [/step] [step:Combine the two implications] The first step proves that boundedness of $A$ implies $\operatorname{diam}(A)<\infty$. The second step proves that $\operatorname{diam}(A)<\infty$ implies boundedness of $A$. Therefore $A$ is bounded if and only if \begin{align*} \operatorname{diam}(A)<\infty. \end{align*} [/step]