[proofplan] We prove the two implications directly from the definitions. If $A$ is [totally bounded](/page/Totally%20Bounded), then a finite cover by sufficiently small open balls gives a finite cover by sets whose diameters are smaller than the prescribed $\varepsilon$. Conversely, a finite cover by sets of diameter less than $\varepsilon$ can be converted into a finite $\varepsilon$-ball cover by choosing one representative point of $A$ from each covering set that actually meets $A$. [/proofplan]

[step:Convert a finite small-ball cover into a finite small-diameter cover] Assume that $A$ is totally bounded in $M$, and let $\varepsilon > 0$. Define \begin{align*} r := \frac{\varepsilon}{3}. \end{align*} Since $r > 0$ and $A$ is totally bounded, there exist an integer $N \geq 0$ and points $x_1,\dots,x_N \in M$ such that \begin{align*} A \subset \bigcup_{i=1}^{N} B(x_i,r), \end{align*} where \begin{align*} B(x_i,r) := \{z \in M : d(z,x_i) < r\}. \end{align*} For each index $i \in \{1,\dots,N\}$, define $E_i := B(x_i,r)$. Then $\{E_1,\dots,E_N\}$ is a finite cover of $A$ by subsets of $M$. If $u,v \in E_i$, then the triangle inequality gives \begin{align*} d(u,v) \leq d(u,x_i) + d(x_i,v) < r + r = \frac{2\varepsilon}{3} < \varepsilon. \end{align*} Taking the supremum over all $u,v \in E_i$ gives \begin{align*} \operatorname{diam}_d(E_i) \leq \frac{2\varepsilon}{3} < \varepsilon \end{align*} for every $i \in \{1,\dots,N\}$. Hence $A$ admits a finite cover by subsets of diameter less than $\varepsilon$. [/step]

[step:Choose representatives from the small-diameter cover]Assume conversely that for every $\varepsilon > 0$ there exists a finite cover of $A$ by subsets of diameter less than $\varepsilon$. Let $\varepsilon > 0$. By hypothesis, there is a finite family $\mathcal{E}$ of subsets of $M$ such that \begin{align*} A \subset \bigcup_{E \in \mathcal{E}} E \end{align*} and $\operatorname{diam}_d(E) < \varepsilon$ for every nonempty $E \in \mathcal{E}$. Define the subfamily $\mathcal{F} := \{E \in \mathcal{E} : E \cap A \neq \varnothing\}$. Since $\mathcal{E}$ is finite, $\mathcal{F}$ is finite. For each $E \in \mathcal{F}$, choose one point $a_E \in E \cap A$.[/step]

[guided]We are proving the converse implication for a fixed number $\varepsilon > 0$. The hypothesis gives a finite family $\mathcal{E}$ of subsets of $M$ such that \begin{align*} A \subset \bigcup_{E \in \mathcal{E}} E \end{align*} and every nonempty $E \in \mathcal{E}$ satisfies $\operatorname{diam}_d(E) < \varepsilon$. The small-diameter cover may contain sets that do not meet $A$. Those sets play no role in covering $A$, and they cannot provide a representative point from $A$. We therefore remove them by defining \begin{align*} \mathcal{F} := \{E \in \mathcal{E} : E \cap A \neq \varnothing\}. \end{align*} This family is finite because it is a subfamily of the finite family $\mathcal{E}$. For every $E \in \mathcal{F}$, the intersection $E \cap A$ is nonempty by definition of $\mathcal{F}$. Hence we may choose a point $a_E \in E \cap A$. These points will become the centers of the finitely many $\varepsilon$-balls. To see that they cover $A$, let $x \in A$. Since $\mathcal{E}$ covers $A$, there exists $E \in \mathcal{E}$ such that $x \in E$. Because $x \in A$, this same set satisfies $E \cap A \neq \varnothing$, so $E \in \mathcal{F}$. The chosen point $a_E$ lies in $E \cap A$, and therefore both $x$ and $a_E$ lie in the nonempty set $E$. By the small-diameter hypothesis on $E$, we have \begin{align*} d(x,a_E) \leq \operatorname{diam}_d(E) < \varepsilon. \end{align*} Thus $x \in B(a_E,\varepsilon)$. Since $x \in A$ was arbitrary, the finite family of balls $\{B(a_E,\varepsilon) : E \in \mathcal{F}\}$ covers $A$. This proves the [total boundedness](/page/Total%20Boundedness) condition for the chosen $\varepsilon > 0$.[/guided]

[step:Show the representative balls cover $A$] We claim that \begin{align*} A \subset \bigcup_{E \in \mathcal{F}} B(a_E,\varepsilon). \end{align*} Let $x \in A$. Since $\mathcal{E}$ covers $A$, there exists $E \in \mathcal{E}$ such that $x \in E$. Because $x \in A$ also, we have $x \in E \cap A$, so $E \in \mathcal{F}$. The point $a_E$ was chosen in $E \cap A$, hence both $x$ and $a_E$ belong to $E$. Since $\operatorname{diam}_d(E) < \varepsilon$, we obtain \begin{align*} d(x,a_E) \leq \operatorname{diam}_d(E) < \varepsilon. \end{align*} Therefore $x \in B(a_E,\varepsilon)$. Since $x \in A$ was arbitrary, the displayed finite union of open $\varepsilon$-balls covers $A$. Because $\mathcal{F}$ is finite, enumerate it as $\mathcal{F} = \{F_1,\dots,F_N\}$ for some integer $N \geq 0$. For each index $j \in \{1,\dots,N\}$, define $y_j := a_{F_j} \in M$. The preceding containment becomes \begin{align*} A \subset \bigcup_{j=1}^{N} B(y_j,\varepsilon). \end{align*} Thus for the given $\varepsilon > 0$, the set $A$ is covered by finitely many open balls of radius $\varepsilon$. Since $\varepsilon > 0$ was arbitrary, $A$ is totally bounded in $M$. [/step]