I'm trying to prove this result.
Theorem: Let $(E, d)$ be a totally bounded metric space, $R >0$, and $\mathcal F$ the set of all real-valued $1$-Lipschitz functions $f$ on $E$ such that $\sup_{x\in E} |f(x)| \le R$. Then $\mathcal F$ is totally bounded w.r.t. the metric of uniform convergence.
Could you have a check if my proof is fine or contains subtle logical mistakes (that I could not recognize)?
My attempt: Fix $\varepsilon>0$ and let $x_1, \ldots, x_n \in E$ be centers of open balls with radius $\varepsilon$ that cover $E$. We define a map $$ \varphi: \mathcal F \to \mathbb R^n, f \mapsto \left ( \varepsilon \left \lfloor \frac{f(x_1)}{\varepsilon} \right \rfloor, \ldots, \varepsilon \left \lfloor \frac{f(x_n)}{\varepsilon} \right \rfloor \right ). $$
Clearly, $$ |f(x_i) - (\varphi(f))_i| < \varepsilon \quad \forall f \in \mathcal F, \forall i = 1, \ldots,n. $$
Here $(\varphi(f))_i$ is the $i$-th coordinate of the $n$-tuple $\varphi (f)$. Because $$ |f (x)| \le R \quad \forall f \in \mathcal F, \forall x \in E, $$ we have $\mathcal K :=\operatorname{im} (\varphi)$ is finite. There is an "inverse" $\psi:\mathcal K \to \mathcal F$ of $\varphi$ such that $\varphi \circ \psi (k) = k$. Clearly, $\mathcal C :=\operatorname{im} (\psi)$ is finite. We claim that $\mathcal C$ is a collection of centers of open balls with radius $\varepsilon$ that cover $\mathcal F$.
Fix $f \in \mathcal F$ and let $g := \psi \circ \varphi (f) \in \mathcal C$. It follows that $\varphi (f) = \varphi (g) =: k$. For $x \in E$, we pick some $i \in \{ 1, \ldots, n\}$ such that $x \in B(x_i, \varepsilon)$. We have $$ \begin{align} |f(x)-g(x)| &\le |f(x)-f(x_i)| + |f(x_i)-g(x_i)| +|g(x_i)-g(x)| \\ &\le \varepsilon + \varepsilon + |f(x_i) - k_i| + |g(x_i) - k_i| \\ &\le 2 \varepsilon + \varepsilon + \varepsilon \\ &=4 \varepsilon. \end{align} $$
This completes the proof.