[proofplan] The proof converts local control from equicontinuity into global control by compactness. We first use equicontinuity with tolerance $1$ to build, around each point of $K$, an open ball on which every function $f_k$ differs from its value at the center by at most $1$. Compactness gives finitely many such balls covering $K$, and pointwise boundedness gives a finite bound at their finitely many centers. The triangle inequality then transfers those finitely many pointwise bounds to a single uniform bound on all of $K$. [/proofplan]

[step:Handle the empty compact space separately] Let $\mathbb{F}\in\{\mathbb{R},\mathbb{C}\}$ denote the common scalar codomain of the functions. If $K=\varnothing$, then the bound $M=0$ satisfies $|f_k(x)|\le M$ for every $x\in K$ and every $k\in\mathbb{N}$, because there are no points $x\in K$. Hence assume from now on that $K\ne\varnothing$. [/step]

[step:Use equicontinuity with tolerance $1$ to build a compact open cover]Fix $a\in K$. By equicontinuity of the sequence $(f_k)_{k=1}^{\infty}$ at the point $a$ with tolerance $1$, there exists $\delta_a>0$ such that for every $x\in K$ and every $k\in\mathbb{N}$, \begin{align*} d(x,a)<\delta_a \implies |f_k(x)-f_k(a)|<1. \end{align*} Define the open ball in the metric subspace $K$ centered at $a$ with radius $\delta_a$ by \begin{align*} U_a:=\{x\in K:d(x,a)<\delta_a\}. \end{align*} Then $(U_a)_{a\in K}$ is an [open cover](/page/Open%20Cover) of $K$, since each $a\in K$ belongs to $U_a$. Because $K$ is compact, this open cover has a finite subcover. Thus there exist points $a_1,\dots,a_N\in K$, with $N\in\mathbb{N}$, such that \begin{align*} K=\bigcup_{i=1}^{N}U_{a_i}. \end{align*}[/step]

[guided]We want to turn pointwise boundedness into a uniform bound. Pointwise boundedness alone gives a possibly different bound at every point, so the central task is to reduce the problem to finitely many points. Equicontinuity provides the local bridge. Fix a point $a\in K$. Equicontinuity of the sequence $(f_k)_{k=1}^{\infty}$ means that, for every tolerance $\varepsilon>0$, there is a radius depending on $a$ and $\varepsilon$, but not depending on $k$, such that all functions $f_k$ vary by less than $\varepsilon$ inside that radius. We choose the specific tolerance $\varepsilon=1$. Therefore there exists $\delta_a>0$ such that for every $x\in K$ and every $k\in\mathbb{N}$, \begin{align*} d(x,a)<\delta_a \implies |f_k(x)-f_k(a)|<1. \end{align*} Define \begin{align*} U_a:=\{x\in K:d(x,a)<\delta_a\}. \end{align*} This is an open ball in the [metric space](/page/Metric%20Space) $(K,d)$. The family $(U_a)_{a\in K}$ covers $K$: indeed, if $a\in K$, then $d(a,a)=0<\delta_a$, so $a\in U_a$. Now compactness is used exactly once. Since $(U_a)_{a\in K}$ is an open cover of the compact metric space $K$, there are finitely many centers $a_1,\dots,a_N\in K$, where $N\in\mathbb{N}$, such that \begin{align*} K=\bigcup_{i=1}^{N}U_{a_i}. \end{align*} The finite list of centers is what allows the pointwise bounds at individual points to be combined into one numerical bound.[/guided]

[step:Combine the pointwise bounds at the finitely many centers] For each index $i\in\{1,\dots,N\}$, pointwise boundedness at the point $a_i$ gives a number $B_i<\infty$ such that \begin{align*} |f_k(a_i)|\le B_i \end{align*} for every $k\in\mathbb{N}$. Define \begin{align*} M:=1+\max_{1\le i\le N}B_i. \end{align*} The maximum exists and is finite because it is taken over the finite set $\{B_1,\dots,B_N\}$. [/step]

[step:Transfer the finite center bounds to every point of $K$] Let $x\in K$ and $k\in\mathbb{N}$. Since $K=\bigcup_{i=1}^{N}U_{a_i}$, there exists $i\in\{1,\dots,N\}$ such that $x\in U_{a_i}$. By the definition of $U_{a_i}$, we have $d(x,a_i)<\delta_{a_i}$, so the radius chosen from equicontinuity gives \begin{align*} |f_k(x)-f_k(a_i)|<1. \end{align*} Using the triangle inequality in $\mathbb{F}$, we obtain \begin{align*} |f_k(x)|\le |f_k(x)-f_k(a_i)|+|f_k(a_i)|. \end{align*} Combining the equicontinuity estimate with the center bound gives \begin{align*} |f_k(x)|<1+B_i\le 1+\max_{1\le j\le N}B_j=M. \end{align*} Thus $|f_k(x)|\le M$ for every $x\in K$ and every $k\in\mathbb{N}$. Since $M<\infty$, the sequence $(f_k)_{k=1}^{\infty}$ is uniformly bounded on $K$. [/step]