[proofplan] We prove the equivalences in a cycle $(1) \Rightarrow (2) \Rightarrow (3) \Rightarrow (4) \Rightarrow (1)$. The implication $(1) \Rightarrow (2)$ is immediate from the definition. For $(2) \Rightarrow (3)$, we show that any $\varepsilon$-net for $A$ is a $2\varepsilon$-net for $\overline{A}$ by the triangle inequality. For $(3) \Rightarrow (4)$, we observe that finite sets (which are totally bounded, since any finite set is its own $\delta$-net for every $\delta > 0$) can always serve as $\varepsilon$-nets. For $(4) \Rightarrow (1)$, we use a finite $\varepsilon/2$-net for the totally bounded $\varepsilon/2$-net to build an $\varepsilon$-net for $A$. [/proofplan]

[step:$(1) \Rightarrow (2)$: An $\varepsilon$-net in open balls is an $\varepsilon$-net in closed balls] Assume $A$ is totally bounded. Let $\varepsilon > 0$. By definition, there exists a finite set $F = \{x_1, \ldots, x_N\} \subset M$ with $A \subset \bigcup_{i=1}^N B(x_i, \varepsilon)$. Since $B(x_i, \varepsilon) \subset \overline{B}_\varepsilon(\{x_i\}) := \{y \in M : d(y, x_i) \le \varepsilon\}$ for each $i$, we have $A \subset \overline{B}_\varepsilon(F)$. This is condition (2). [/step]

[step:$(2) \Rightarrow (3)$: Extend the covering from $A$ to $\overline{A}$ via the triangle inequality]Assume (2). Let $\varepsilon > 0$. Applying (2) with $\varepsilon/2$, there exists a finite set $F \subset M$ with $A \subset \overline{B}_{\varepsilon/2}(F)$, meaning every $a \in A$ satisfies $\operatorname{dist}(a, F) \le \varepsilon/2$. Let $y \in \overline{A}$. Then there exists a sequence $\{a_k\}_{k=1}^\infty$ in $A$ with $d(a_k, y) \to 0$. In particular, there exists $a \in A$ with $d(a, y) < \varepsilon/2$. Since $\operatorname{dist}(a, F) \le \varepsilon/2$, there exists $x \in F$ with $d(a, x) \le \varepsilon/2$. By the triangle inequality, \begin{align*} d(y, x) \le d(y, a) + d(a, x) < \frac{\varepsilon}{2} + \frac{\varepsilon}{2} = \varepsilon. \end{align*} Therefore $y \in B(x, \varepsilon) \subset \bigcup_{z \in F} B(z, \varepsilon)$, so $\overline{A} \subset \bigcup_{z \in F} B(z, \varepsilon)$. Since $F$ is finite, $\overline{A}$ is totally bounded.[/step]

[guided]Assume (2): for every $\varepsilon > 0$, there exists a finite $F \subset M$ with $A \subset \overline{B}_\varepsilon(F)$. We must show that $\overline{A}$ is totally bounded. The idea is that points in $\overline{A}$ are arbitrarily close to points in $A$, and points in $A$ are within distance $\varepsilon/2$ of $F$. Using the triangle inequality with the "budget" split $\varepsilon = \varepsilon/2 + \varepsilon/2$ converts a covering of $A$ at scale $\varepsilon/2$ into a covering of $\overline{A}$ at scale $\varepsilon$. Fix $\varepsilon > 0$. Apply (2) with parameter $\varepsilon/2$ to obtain a finite set $F \subset M$ with $A \subset \overline{B}_{\varepsilon/2}(F)$. Take any $y \in \overline{A}$. By the definition of closure, for every $\delta > 0$, the ball $B(y, \delta)$ intersects $A$. Choosing $\delta = \varepsilon/2$, there exists $a \in A$ with $d(a, y) < \varepsilon/2$. Since $a \in A \subset \overline{B}_{\varepsilon/2}(F)$, there exists $x \in F$ with $d(a, x) \le \varepsilon/2$. The triangle inequality gives \begin{align*} d(y, x) \le d(y, a) + d(a, x) < \frac{\varepsilon}{2} + \frac{\varepsilon}{2} = \varepsilon. \end{align*} Thus $y \in B(x, \varepsilon)$, so $\overline{A} \subset \bigcup_{z \in F} B(z, \varepsilon)$. Since $\varepsilon > 0$ was arbitrary and $F$ is finite, $\overline{A}$ is totally bounded.[/guided]

[step:$(3) \Rightarrow (4)$: Finite $\varepsilon$-nets are automatically totally bounded] Assume $\overline{A}$ is totally bounded. Since $A \subset \overline{A}$, the set $A$ is itself totally bounded: for every $\varepsilon > 0$, any $\varepsilon$-net for $\overline{A}$ is also an $\varepsilon$-net for $A$ (as $A \subset \overline{A} \subset \bigcup_{i=1}^N B(x_i, \varepsilon)$). For condition (4), we must show that the $\varepsilon$-net can be chosen to be totally bounded. The $\varepsilon$-nets produced by the definition are finite sets, and every finite subset of a metric space is totally bounded: given $\delta > 0$ and a finite set $F = \{x_1, \ldots, x_N\}$, the set $F$ itself is a $\delta$-net for $F$ (each $x_i \in B(x_i, \delta)$). Thus condition (4) is satisfied. [/step]

[step:$(4) \Rightarrow (1)$: Compose two finite coverings to recover total boundedness of $A$]Assume (4): for every $\varepsilon > 0$, there exists a totally bounded set $G_\varepsilon \subset M$ that is an $\varepsilon$-net for $A$, i.e., $A \subset \bigcup_{g \in G_\varepsilon} B(g, \varepsilon)$. Fix $\varepsilon > 0$. Apply (4) with $\varepsilon/2$ to obtain a totally bounded set $G := G_{\varepsilon/2}$ with $A \subset \bigcup_{g \in G} B(g, \varepsilon/2)$. Since $G$ is totally bounded, there exists a finite set $F = \{x_1, \ldots, x_N\} \subset M$ with $G \subset \bigcup_{i=1}^N B(x_i, \varepsilon/2)$. For any $a \in A$, there exists $g \in G$ with $d(a, g) < \varepsilon/2$, and there exists $x_i \in F$ with $d(g, x_i) < \varepsilon/2$. By the triangle inequality, \begin{align*} d(a, x_i) \le d(a, g) + d(g, x_i) < \frac{\varepsilon}{2} + \frac{\varepsilon}{2} = \varepsilon. \end{align*} Therefore $A \subset \bigcup_{i=1}^N B(x_i, \varepsilon)$, and $F$ is a finite $\varepsilon$-net for $A$. Since $\varepsilon > 0$ was arbitrary, $A$ is totally bounded.[/step]

[guided]Assume (4): for every $\varepsilon > 0$, there exists a totally bounded $\varepsilon$-net $G_\varepsilon$ for $A$. The strategy is to compose two coverings: first cover $A$ by balls centred at points of $G_\varepsilon$, then cover $G_\varepsilon$ (which is totally bounded) by finitely many balls. The triangle inequality composes the two radii. Fix $\varepsilon > 0$. Apply (4) with parameter $\varepsilon/2$ to obtain a totally bounded set $G \subset M$ satisfying $A \subset \bigcup_{g \in G} B(g, \varepsilon/2)$. Now use the total boundedness of $G$: there exists a finite set $F = \{x_1, \ldots, x_N\} \subset M$ with $G \subset \bigcup_{i=1}^N B(x_i, \varepsilon/2)$. We claim $F$ is a finite $\varepsilon$-net for $A$. Let $a \in A$. Since $A \subset \bigcup_{g \in G} B(g, \varepsilon/2)$, there exists $g \in G$ with $d(a, g) < \varepsilon/2$. Since $G \subset \bigcup_{i=1}^N B(x_i, \varepsilon/2)$, there exists $x_i \in F$ with $d(g, x_i) < \varepsilon/2$. The triangle inequality yields \begin{align*} d(a, x_i) \le d(a, g) + d(g, x_i) < \frac{\varepsilon}{2} + \frac{\varepsilon}{2} = \varepsilon. \end{align*} Thus $a \in B(x_i, \varepsilon)$, and since $a \in A$ was arbitrary, $A \subset \bigcup_{i=1}^N B(x_i, \varepsilon)$. This "covering of a covering" argument is a recurring motif in metric space theory. The key is the triangle inequality's ability to compose two $\varepsilon/2$-approximations into a single $\varepsilon$-approximation, and the observation that total boundedness of the intermediate set $G$ reduces the (potentially infinite) first covering to a finite one.[/guided]