[proofplan] We prove [total boundedness](/page/Total%20Boundedness) of $f(A)$ directly from the finite-cover definition. Given a target radius $\varepsilon > 0$ in $N$, [uniform continuity](/page/Uniform%20Continuity) supplies a single radius $\delta > 0$ in $M$ that works for all points of $M$. Total boundedness of $A$ then gives a finite $\delta/2$-cover of $A$; choosing one representative point of $A$ from each nonempty covering ball produces finitely many centers whose images cover $f(A)$ by $\varepsilon$-balls. [/proofplan] [step:Dispose of the empty subset] If $A = \varnothing$, then $f(A)=\varnothing$. For every $\varepsilon > 0$, the finite set $\varnothing \subset N$ is an $\varepsilon$-net for $f(A)$, because there is no point of $f(A)$ to check. Hence $f(A)$ is [totally bounded](/page/Totally%20Bounded) in $N$ in this case. We may therefore assume for the rest of the proof that $A \neq \varnothing$. [/step] [step:Choose a uniform continuity scale for the target radius] Let $\varepsilon > 0$ be arbitrary. Since $f: M \to N$ is uniformly continuous, there exists $\delta > 0$ such that for all $x,y \in M$, \begin{align*} d_M(x,y) < \delta \implies d_N(f(x),f(y)) < \varepsilon. \end{align*} [/step] [step:Replace an ambient finite cover by representatives from $A$] Since $A$ is totally bounded in $M$ and $\delta/2 > 0$, there exist a positive integer $k$ and points $x_1,\dots,x_k \in M$ such that \begin{align*} A \subset \bigcup_{i=1}^{k} B_M(x_i,\delta/2), \end{align*} where $B_M(x_i,\delta/2) := \{x \in M : d_M(x,x_i) < \delta/2\}$. Define the finite index set \begin{align*} J := \{i \in \{1,\dots,k\} : A \cap B_M(x_i,\delta/2) \neq \varnothing\}. \end{align*} For each $i \in J$, choose a point $a_i \in A \cap B_M(x_i,\delta/2)$. Define the finite subset $F \subset f(A)$ by \begin{align*} F := \{f(a_i) : i \in J\}. \end{align*} [guided] The finite cover of $A$ obtained from total boundedness may have centers $x_i$ in the ambient space $M$, not necessarily in $A$. This is harmless, but to ensure that the eventual centers in $N$ lie in $f(A)$, we choose representatives from $A$. Since $\delta/2 > 0$ and $A$ is totally bounded in $M$, there are finitely many points $x_1,\dots,x_k \in M$ such that \begin{align*} A \subset \bigcup_{i=1}^{k} B_M(x_i,\delta/2). \end{align*} Some of these balls may not meet $A$; those balls do not contribute to covering points of $A$. We therefore define \begin{align*} J := \{i \in \{1,\dots,k\} : A \cap B_M(x_i,\delta/2) \neq \varnothing\}. \end{align*} For every $i \in J$, the set $A \cap B_M(x_i,\delta/2)$ is nonempty by definition of $J$, so we may choose a representative point $a_i \in A \cap B_M(x_i,\delta/2)$. The set \begin{align*} F := \{f(a_i) : i \in J\} \end{align*} is finite because $J$ is finite, and it is a subset of $f(A)$ because every representative $a_i$ lies in $A$. [/guided] [/step] [step:Show the finitely many image representatives cover $f(A)$] We claim that \begin{align*} f(A) \subset \bigcup_{z \in F} B_N(z,\varepsilon), \end{align*} where $B_N(z,\varepsilon) := \{w \in N : d_N(w,z) < \varepsilon\}$. Let $w \in f(A)$. Then there exists $a \in A$ such that $w = f(a)$. Since the balls $B_M(x_i,\delta/2)$ cover $A$, there exists $i \in \{1,\dots,k\}$ such that $a \in B_M(x_i,\delta/2)$. This index belongs to $J$, and the representative $a_i \in A \cap B_M(x_i,\delta/2)$ satisfies \begin{align*} d_M(a,a_i) \le d_M(a,x_i) + d_M(x_i,a_i) < \delta/2 + \delta/2 = \delta. \end{align*} By the choice of $\delta$ from uniform continuity, \begin{align*} d_N(w,f(a_i)) = d_N(f(a),f(a_i)) < \varepsilon. \end{align*} Thus $w \in B_N(f(a_i),\varepsilon)$, and $f(a_i) \in F$. Hence the displayed finite union covers $f(A)$. [/step] [step:Conclude total boundedness of the image] For the arbitrary radius $\varepsilon > 0$, we have produced a finite set $F \subset f(A)$ such that \begin{align*} f(A) \subset \bigcup_{z \in F} B_N(z,\varepsilon). \end{align*} Therefore $f(A)$ is totally bounded in $N$. [/step]