By Corollary (complete [metric space](/page/Metric%20Space): precompact $\iff$ totally bounded), it suffices to prove: $F$ is totally bounded $\iff$ $F$ is bounded and equicontinuous. **Step 1: ($\Rightarrow$) Total boundedness implies bounded and equicontinuous.** *Bounded:* Take $\varepsilon = 1$ in the definition of total boundedness. There exist $f_1, \ldots, f_m \in C(K)$ such that every $f \in F$ satisfies $\|f - f_i\|_{C(K)} < 1$ for some $i$. Hence $\|f\|_{C(K)} \le 1 + \max_i \|f_i\|_{C(K)}$ for all $f \in F$. *Equicontinuous:* Let $\varepsilon > 0$. Choose a finite $\varepsilon$-net $\{f_1, \ldots, f_m\}$ for $F$. Fix $x \in K$. Since each $f_i$ is [continuous](/page/Continuity), there exists an open neighbourhood $U_i \ni x$ with $|f_i(y) - f_i(x)| < \varepsilon$ for all $y \in U_i$. Set $U = \bigcap_{i=1}^m U_i$, which is an open neighbourhood of $x$. For any $f \in F$, choose $f_i$ with $\|f - f_i\|_{C(K)} < \varepsilon$. Then for $y \in U$: \begin{align*} |f(y) - f(x)| \le |f(y) - f_i(y)| + |f_i(y) - f_i(x)| + |f_i(x) - f(x)| < 3\varepsilon. \end{align*} **Step 2: ($\Leftarrow$) Bounded and equicontinuous implies totally bounded.** Let $\varepsilon > 0$. By equicontinuity, for each $x \in K$ there exists an open $U_x \ni x$ with $|f(y) - f(x)| < \varepsilon$ for all $y \in U_x$ and all $f \in F$. Since $K$ is compact, finitely many of these cover $K$: $K \subset \bigcup_{i=1}^n U_{x_i}$. Consider the restriction of $F$ to $\{x_1, \ldots, x_n\}$, viewed as a subset of $\ell_n^\infty$. Since $F$ is bounded, this restricted set is a bounded subset of a finite-dimensional space, hence totally bounded. So there exist $f_1, \ldots, f_m \in F$ forming an $\varepsilon$-net for the restricted set: for every $f \in F$, some $f_j$ satisfies $\max_{1 \le i \le n} |f(x_i) - f_j(x_i)| < \varepsilon$. We claim $\{f_1, \ldots, f_m\}$ is a $3\varepsilon$-net for $F$ in $C(K)$. For any $f \in F$ and $y \in K$, choose $x_i$ with $y \in U_{x_i}$, and $f_j$ with $\max_i |f(x_i) - f_j(x_i)| < \varepsilon$: \begin{align*} |f(y) - f_j(y)| \le |f(y) - f(x_i)| + |f(x_i) - f_j(x_i)| + |f_j(x_i) - f_j(y)| < \varepsilon + \varepsilon + \varepsilon = 3\varepsilon \end{align*} where the first and third terms use equicontinuity of $F$ (both $f$ and $f_j$ belong to $F$). Since $y \in K$ was arbitrary, $\|f - f_j\|_{C(K)} \le 3\varepsilon$.