The Arzela-Ascoli Theorem

Step 1 — A countable dense set. Since $K$ is a compact metric space, it is separable. Choose a countable dense subset $D=\{x_1,x_2,x_3,\dots\}\subseteq K$.

Step 2 — Diagonal extraction of pointwise convergence on $D$. At $x_1$, the sequence $\{f_n(x_1)\}$ lies in a bounded subset of $\mathbb{R}^m$ (by uniform boundedness), hence by Bolzano–Weierstrass has a convergent subsequence $\{f_{n}^{(1)}\}$. From it, the values at $x_2$ are bounded, giving a further subsequence $\{f_{n}^{(2)}\}$ converging at $x_2$ (and still at $x_1$). Iterate. Define the diagonal sequence

Derivation hint: $\{g_n\}_{n\ge k}$ is, after the $k$-th term, a subsequence of $\{f_n^{(k)}\}$, so it converges at $x_k$ for every $k$ — the classic Cantor diagonal trick. Thus $g_n(x)$ converges for every $x\in D$.

Step 3 — Cauchy estimate via equicontinuity (the spreading). Fix $\varepsilon>0$. By equicontinuity (uniform, since $K$ compact) pick $\delta>0$ so that

Derivation hint: Use Definition [2] with the single $\delta$ valid for the whole family $\{g_n\}\subseteq\mathcal F$. By compactness, $K$ is covered by finitely many balls $B(y_j,\delta)$, $j=1,\dots,N$, with $y_j\in D$ (density lets us put centers in $D$).

Step 4 — Uniform Cauchy. For each $j$, $\{g_n(y_j)\}$ converges, hence is Cauchy: choose $M$ such that $n,p\ge M \implies \|g_n(y_j)-g_p(y_j)\| < \varepsilon/3$ for all $j=1,\dots,N$. For arbitrary $x\in K$, pick $y_j$ with $d(x,y_j)<\delta$. Then for $n,p\ge M$:

Derivation hint: Triangle inequality; first and third terms controlled by [4], middle by Step 4's pointwise Cauchy choice — and $j$ depends on $x$ but the bound $M$ does not. Therefore $\{g_n\}$ is uniformly Cauchy. Since $(C(K),\|\cdot\|_\infty)$ is complete, $g_n \to g$ uniformly, and $g\in C(K)$ (uniform limit of continuous functions is continuous). $\qquad\blacksquare$

Suppose $\overline{\mathcal{F}}$ is compact in $C(K)$.

Bounded. A compact subset of a metric space is bounded, so $\sup_{f\in\mathcal F}\|f\|_\infty<\infty$ — uniform boundedness.

Equicontinuous. Compactness $\Rightarrow$ total boundedness: given $\varepsilon>0$ there exist $h_1,\dots,h_p\in\mathcal F$ forming an $\varepsilon/3$-net, i.e. every $f\in\mathcal F$ satisfies $\|f-h_k\|_\infty<\varepsilon/3$ for some $k$. Each $h_k$ is uniformly continuous on compact $K$, so there is $\delta_k$ with $d(x,x')<\delta_k \Rightarrow \|h_k(x)-h_k(x')\|<\varepsilon/3$. Let $\delta=\min_k\delta_k$. Then for any $f$ and $d(x,x')<\delta$:

Derivation hint: The finitely many $h_k$ "anchor" the whole family; total boundedness converts a continuity statement for each anchor into equicontinuity for all of $\mathcal F$. $\qquad\blacksquare$

Assume $\mathcal F$ pointwise equicontinuous on compact $K$; fix $\varepsilon>0$. For each $x\in K$ pick $\delta_x>0$ so that $d(z,x)<\delta_x \Rightarrow \|f(z)-f(x)\|<\varepsilon/2$ for all $f$. The balls $\{B(x,\delta_x/2)\}$ cover $K$; extract a finite subcover with centers $x_1,\dots,x_N$ and set $\delta=\tfrac12\min_i\delta_{x_i}$. If $d(x,x')<\delta$, then $x\in B(x_i,\delta_{x_i}/2)$ for some $i$, and $d(x',x_i)\le d(x',x)+d(x,x_i)<\delta+\delta_{x_i}/2\le\delta_{x_i}$, so both lie within $\delta_{x_i}$ of $x_i$; hence

Derivation hint: A Lebesgue-number / finite-subcover argument; the same trick that promotes continuity to uniform continuity on a compact set. $\qquad\blacksquare$