[proofplan] We handle the empty set separately, since boundedness of the empty set is vacuous. For a nonempty totally [bounded set](/page/Bounded%20Set), apply [total boundedness](/page/Total%20Boundedness) with radius $1$ to cover $A$ by finitely many metric balls. Choosing one centre as a reference point, the triangle inequality bounds the distance from that reference point to every element of $A$ by the maximum of finitely many centre-to-centre distances plus $1$. [/proofplan] [step:Dispose of the empty set case] If $A=\varnothing$, then $A$ is bounded by the meaning of boundedness stated in the theorem. Hence it remains to prove the claim under the assumption that $A\neq\varnothing$. [/step] [step:Extract a finite unit ball cover from total boundedness] Assume $A\neq\varnothing$. Since $A$ is totally bounded, applying the definition with $\varepsilon=1$ gives points $x_1,\dots,x_m\in X$, with $m\in\mathbb N$, such that \begin{align*} A\subset \bigcup_{i=1}^{m} B_d(x_i,1). \end{align*} Thus, for every $a\in A$, there exists an index $i\in\{1,\dots,m\}$ such that $d(a,x_i)<1$. [guided] The definition of total boundedness says that every positive radius admits a finite ball cover of the set. We choose the particular radius $\varepsilon=1$ because boundedness only requires some finite radius, not an optimal one. Since $A$ is totally bounded, there exist finitely many points $x_1,\dots,x_m\in X$, with $m\in\mathbb N$, such that \begin{align*} A\subset \bigcup_{i=1}^{m} B_d(x_i,1). \end{align*} This means precisely that every point $a\in A$ lies in at least one of these balls. Therefore, for each $a\in A$, there is an index $i\in\{1,\dots,m\}$ satisfying \begin{align*} d(a,x_i)<1. \end{align*} The finiteness of the list of centres is the key input: it will let us take a maximum of their distances from one fixed reference centre. [/guided] [/step] [step:Choose a reference centre and bound all cover centres from it] Because $m\in\mathbb N$, the finite list of centres contains at least one point. Define the reference point $x_0:=x_1\in X$. Since the set \begin{align*} \{d(x_0,x_i):i\in\{1,\dots,m\}\} \end{align*} is a finite subset of $\mathbb R$, it has a maximum. Define \begin{align*} M:=\max_{1\le i\le m} d(x_0,x_i). \end{align*} Then $M<\infty$. [/step] [step:Use the triangle inequality to place all of $A$ in one ball] Let $a\in A$. Choose $i\in\{1,\dots,m\}$ such that $d(a,x_i)<1$. By symmetry of the metric and the triangle inequality, \begin{align*} d(x_0,a)\le d(x_0,x_i)+d(x_i,a)< M+1. \end{align*} Therefore $a\in B_d(x_0,M+1)$. Since $a\in A$ was arbitrary, \begin{align*} A\subset B_d(x_0,M+1). \end{align*} The radius $M+1$ is finite, so $A$ is bounded. [/step]