[proofplan] We prove the result directly from the finite-ball definition of [total boundedness](/page/Total%20Boundedness). Given a scale $\varepsilon > 0$, we first cover the whole space $M$ by finitely many balls of radius $\varepsilon/2$. For every ball in this finite cover that meets $A$, we choose one point of $A$ inside it; the triangle inequality then shows that the corresponding $\varepsilon$-balls centered in $A$ cover all of $A$. The empty subset is handled separately by the empty finite cover. [/proofplan]

[step:Handle the empty subset by the empty finite cover] For $x \in M$ and $r > 0$, define the open ball \begin{align*} B_M(x,r) := \{y \in M : d(x,y) < r\}. \end{align*} If $A = \varnothing$, then the empty family of balls is a finite cover of $A$. Hence $A$ is [totally bounded](/page/Totally%20Bounded). We therefore assume for the rest of the proof that $A \neq \varnothing$. [/step]

[step:Refine a finite cover of $M$ to centers lying in $A$]Let $\varepsilon > 0$ be arbitrary, and define $\rho := \varepsilon/2$. Since $M$ is totally bounded, there exist an integer $N \in \mathbb{N}$ and points $x_1,\dots,x_N \in M$ such that \begin{align*} M \subset \bigcup_{i=1}^{N} B_M(x_i,\rho). \end{align*} Define the finite index set \begin{align*} I := \{i \in \{1,\dots,N\} : A \cap B_M(x_i,\rho) \neq \varnothing\}. \end{align*} For each $i \in I$, choose a point $a_i \in A \cap B_M(x_i,\rho)$. This choice is over a finite index set. We claim that the finite family $\{B_M(a_i,\varepsilon) : i \in I\}$ covers $A$. Indeed, let $a \in A$. Since $A \subset M$ and the balls $B_M(x_i,\rho)$ cover $M$, there exists $i \in \{1,\dots,N\}$ such that $a \in B_M(x_i,\rho)$. Therefore $A \cap B_M(x_i,\rho) \neq \varnothing$, so $i \in I$. By construction, $a_i \in B_M(x_i,\rho)$, and hence \begin{align*} d(a,a_i) \leq d(a,x_i) + d(x_i,a_i) < \rho + \rho = \varepsilon. \end{align*} Thus $a \in B_M(a_i,\varepsilon)$.[/step]

[guided]The only subtle point is the possible convention about where the centers of a finite cover must lie. Some definitions of a totally bounded subset of $M$ allow the covering balls to have centers anywhere in $M$, while others require the centers to lie in the subset itself. We prove the stronger version with centers in $A$. Fix $\varepsilon > 0$ and set $\rho := \varepsilon/2$. The reason for using $\varepsilon/2$ instead of $\varepsilon$ is that we will move the center of each relevant ball from a point $x_i \in M$ to a nearby point $a_i \in A$. The triangle inequality will then double the radius. Since $M$ is totally bounded, there are finitely many points $x_1,\dots,x_N \in M$ such that \begin{align*} M \subset \bigcup_{i=1}^{N} B_M(x_i,\rho). \end{align*} Not every ball in this cover has to meet $A$, so we keep only the relevant indices. Define \begin{align*} I := \{i \in \{1,\dots,N\} : A \cap B_M(x_i,\rho) \neq \varnothing\}. \end{align*} For each $i \in I$, the set $A \cap B_M(x_i,\rho)$ is nonempty by definition of $I$, so we may choose a point $a_i \in A \cap B_M(x_i,\rho)$. The set $I$ is finite because $I \subset \{1,\dots,N\}$. We now verify that the balls centered at these selected points cover $A$. Let $a \in A$. Since $a \in M$ and the balls $B_M(x_i,\rho)$ cover $M$, there is an index $i \in \{1,\dots,N\}$ with $a \in B_M(x_i,\rho)$. This also shows that $A \cap B_M(x_i,\rho)$ is nonempty, because it contains $a$. Hence $i \in I$, so the point $a_i \in A \cap B_M(x_i,\rho)$ has been chosen. Now both $a$ and $a_i$ lie within distance $\rho$ of $x_i$. Applying the triangle inequality in the [metric space](/page/Metric%20Space) $(M,d)$ gives \begin{align*} d(a,a_i) \leq d(a,x_i) + d(x_i,a_i) < \rho + \rho = \varepsilon. \end{align*} Therefore $a \in B_M(a_i,\varepsilon)$. Since $a \in A$ was arbitrary, the finite family $\{B_M(a_i,\varepsilon) : i \in I\}$ covers all of $A$.[/guided]

[step:Conclude total boundedness of the subset] We have shown that for every $\varepsilon > 0$ there is a finite set of points $\{a_i : i \in I\} \subset A$ such that \begin{align*} A \subset \bigcup_{i \in I} B_M(a_i,\varepsilon). \end{align*} This is precisely the finite-ball condition for $A$ to be totally bounded with the metric induced by $d$. Hence every subset $A \subset M$ is totally bounded in $M$. [/step]