[proofplan] The proof is a [sequential compactness](/page/Sequential%20Compactness) argument. Boundedness in $C^{0,\gamma}(K;\mathbb R^m)$ gives uniform boundedness and a common modulus of continuity, so compactness of $K$ lets us extract a uniformly convergent subsequence by the usual finite-net diagonal argument underlying Arzelà-Ascoli. The uniform limit inherits the $\gamma$-Hölder bound. Finally, an interpolation estimate between the uniform norm and the $\gamma$-Hölder seminorm upgrades [uniform convergence](/page/Uniform%20Convergence) to convergence in the weaker $C^{0,\alpha}$ norm. [/proofplan]

[step:Reduce to a nonempty compact domain and fix the Hölder bounds] If $K=\varnothing$, then there is exactly one map $K\to\mathbb R^m$, so every sequence is eventually constant as a sequence in both Hölder spaces, and the conclusion follows. Hence assume $K\ne\varnothing$. For a bounded map $h:K\to\mathbb R^m$, define \begin{align*} \|h\|_\infty:=\sup_{x\in K}|h(x)|. \end{align*} For $0<\beta\le1$, define the Hölder seminorm of a map $h:K\to\mathbb R^m$ by \begin{align*} [h]_{C^{0,\beta}(K)}:=\sup_{\substack{x,y\in K\\x\ne y}}\frac{|h(x)-h(y)|}{|x-y|^\beta}, \end{align*} with the convention that the supremum is $0$ when $K$ has at most one point. Since $(f_j)_{j\in\mathbb N}$ is bounded in $C^{0,\gamma}(K;\mathbb R^m)$, define \begin{align*} M:=\sup_{j\in\mathbb N}\|f_j\|_{C^{0,\gamma}(K)}<\infty. \end{align*} Then for every $j\in\mathbb N$ and every $x,y\in K$, \begin{align*} |f_j(x)|\le M \end{align*} and \begin{align*} |f_j(x)-f_j(y)|\le M|x-y|^\gamma. \end{align*} [/step]

[step:Extract a uniformly convergent subsequence by finite nets]Since $K$ is compact in the Euclidean metric inherited from $\mathbb R^n$, it is totally bounded. For each $r\in\mathbb N$, choose a finite set $A_r\subset K$ such that \begin{align*} K\subset\bigcup_{a\in A_r}B(a,1/r). \end{align*} Let $A:=\bigcup_{r=1}^{\infty}A_r$, and enumerate $A$ as a sequence $(a_q)_{q\in\mathbb N}$ after deleting repetitions if necessary. For each fixed $q\in\mathbb N$, the sequence $(f_j(a_q))_{j\in\mathbb N}$ lies in the closed Euclidean ball $\overline{B}(0,M)\subset\mathbb R^m$, which is compact. A diagonal subsequence argument gives a strictly increasing sequence $(j_l)_{l\in\mathbb N}$ such that, for every $q\in\mathbb N$, the sequence $(f_{j_l}(a_q))_{l\in\mathbb N}$ converges in $\mathbb R^m$. We prove that $(f_{j_l})_{l\in\mathbb N}$ is Cauchy in the uniform norm. Let $\varepsilon>0$. If $M=0$, then each $f_j$ is identically zero on $K$, so the assertion is immediate. Assume $M>0$. Choose $r\in\mathbb N$ so large that \begin{align*} M r^{-\gamma}<\frac{\varepsilon}{3}. \end{align*} Since $A_r$ is finite and $(f_{j_l}(a))_{l\in\mathbb N}$ converges for every $a\in A_r$, there exists $L\in\mathbb N$ such that for all $p,q\ge L$ and every $a\in A_r$, \begin{align*} |f_{j_p}(a)-f_{j_q}(a)|<\frac{\varepsilon}{3}. \end{align*} For any $x\in K$, choose $a\in A_r$ with $|x-a|<1/r$. Then, for all $p,q\ge L$, \begin{align*} |f_{j_p}(x)-f_{j_q}(x)|\le |f_{j_p}(x)-f_{j_p}(a)|+|f_{j_p}(a)-f_{j_q}(a)|+|f_{j_q}(a)-f_{j_q}(x)|. \end{align*} The common Hölder bound gives \begin{align*} |f_{j_p}(x)-f_{j_q}(x)|<M r^{-\gamma}+\frac{\varepsilon}{3}+M r^{-\gamma}<\varepsilon. \end{align*} Taking the supremum over $x\in K$ shows that $(f_{j_l})_{l\in\mathbb N}$ is Cauchy in $\|\cdot\|_\infty$. Since $\mathbb R^m$ is complete, there exists a map $f:K\to\mathbb R^m$ such that \begin{align*} \lim_{l\to\infty}\|f_{j_l}-f\|_\infty=0. \end{align*}[/step]

[guided]The compactness of $K$ is used in the concrete finite-net form: for every scale $1/r$, finitely many balls of radius $1/r$ cover $K$. Choose one finite set of centres $A_r\subset K$ for each scale and put $A=\bigcup_{r=1}^{\infty}A_r$. This set is countable and dense enough for the argument because every point of $K$ lies within $1/r$ of some point of $A_r$. At each selected point $a\in A$, the values $f_j(a)$ stay in the compact closed ball $\overline{B}(0,M)\subset\mathbb R^m$. Therefore we can extract subsequences so that the values converge at the first point, then at the second point, and so on. Passing to the diagonal subsequence gives a strictly increasing sequence $(j_l)_{l\in\mathbb N}$ such that, for every fixed $a\in A$, the sequence $(f_{j_l}(a))_{l\in\mathbb N}$ converges in $\mathbb R^m$. Now we upgrade pointwise convergence on the countable net to uniform Cauchy convergence on all of $K$. Let $\varepsilon>0$. If $M=0$, then $\|f_j\|_\infty=0$ for every $j$, so every $f_j$ is the zero map and there is nothing to prove. Assume $M>0$. Choose $r\in\mathbb N$ such that \begin{align*} M r^{-\gamma}<\frac{\varepsilon}{3}. \end{align*} Because $A_r$ is finite and the subsequence converges at each point of $A_r$, there exists $L\in\mathbb N$ such that for every $p,q\ge L$ and every $a\in A_r$, \begin{align*} |f_{j_p}(a)-f_{j_q}(a)|<\frac{\varepsilon}{3}. \end{align*} For an arbitrary point $x\in K$, choose $a\in A_r$ with $|x-a|<1/r$. The triangle inequality compares the two functions at $x$ by moving from $x$ to the nearby net point $a$, comparing the values at $a$, and moving back to $x$: \begin{align*} |f_{j_p}(x)-f_{j_q}(x)|\le |f_{j_p}(x)-f_{j_p}(a)|+|f_{j_p}(a)-f_{j_q}(a)|+|f_{j_q}(a)-f_{j_q}(x)|. \end{align*} The first and third terms are controlled by the common $\gamma$-Hölder estimate: \begin{align*} |f_{j_p}(x)-f_{j_p}(a)|\le M|x-a|^\gamma<M r^{-\gamma} \end{align*} and \begin{align*} |f_{j_q}(a)-f_{j_q}(x)|\le M|a-x|^\gamma<M r^{-\gamma}. \end{align*} Hence, for all $p,q\ge L$, \begin{align*} |f_{j_p}(x)-f_{j_q}(x)|<M r^{-\gamma}+\frac{\varepsilon}{3}+M r^{-\gamma}<\varepsilon. \end{align*} Since the bound is independent of $x$, the subsequence is Cauchy in the uniform norm. Completeness of $\mathbb R^m$ then gives a pointwise limit map $f:K\to\mathbb R^m$, and the uniform Cauchy estimate implies \begin{align*} \lim_{l\to\infty}\|f_{j_l}-f\|_\infty=0. \end{align*}[/guided]

[step:Pass the Hölder bound to the uniform limit] For every $x,y\in K$ and every $l\in\mathbb N$, \begin{align*} |f_{j_l}(x)-f_{j_l}(y)|\le M|x-y|^\gamma. \end{align*} Taking the limit as $l\to\infty$ and using pointwise convergence, we obtain \begin{align*} |f(x)-f(y)|\le M|x-y|^\gamma. \end{align*} Thus $f\in C^{0,\gamma}(K;\mathbb R^m)$ and \begin{align*} [f]_{C^{0,\gamma}(K)}\le M. \end{align*} Since $K$ is compact, its diameter \begin{align*} D:=\operatorname{diam}(K)=\sup_{x,y\in K}|x-y| \end{align*} is finite. If $D=0$, then $K$ has one point and every Hölder seminorm on $K$ is zero. If $D>0$, then for $x\ne y$ in $K$, \begin{align*} \frac{|f(x)-f(y)|}{|x-y|^\alpha}\le M |x-y|^{\gamma-\alpha}\le M D^{\gamma-\alpha}. \end{align*} Therefore $f\in C^{0,\alpha}(K;\mathbb R^m)$. [/step]

[step:Interpolate uniform convergence with the common Hölder bound] Define \begin{align*} u_l:K\to\mathbb R^m,\qquad u_l:=f_{j_l}-f. \end{align*} Then \begin{align*} \lim_{l\to\infty}\|u_l\|_\infty=0. \end{align*} Also, \begin{align*} [u_l]_{C^{0,\gamma}(K)}\le [f_{j_l}]_{C^{0,\gamma}(K)}+[f]_{C^{0,\gamma}(K)}\le 2M. \end{align*} Let \begin{align*} \theta:=\frac{\alpha}{\gamma}\in(0,1). \end{align*} For any $x,y\in K$ with $x\ne y$, we have both estimates \begin{align*} |u_l(x)-u_l(y)|\le 2\|u_l\|_\infty \end{align*} and \begin{align*} |u_l(x)-u_l(y)|\le [u_l]_{C^{0,\gamma}(K)}|x-y|^\gamma. \end{align*} Multiplying the first estimate to the power $1-\theta$ and the second estimate to the power $\theta$ gives \begin{align*} |u_l(x)-u_l(y)|\le (2\|u_l\|_\infty)^{1-\theta}[u_l]_{C^{0,\gamma}(K)}^\theta |x-y|^\alpha. \end{align*} Hence \begin{align*} [u_l]_{C^{0,\alpha}(K)}\le (2\|u_l\|_\infty)^{1-\theta}(2M)^\theta. \end{align*} Since $\|u_l\|_\infty\to0$ and $1-\theta>0$, it follows that \begin{align*} \lim_{l\to\infty}[u_l]_{C^{0,\alpha}(K)}=0. \end{align*} Combining this with uniform convergence yields \begin{align*} \lim_{l\to\infty}\|f_{j_l}-f\|_{C^{0,\alpha}(K)}=0. \end{align*} [/step]

[step:Conclude compactness of the inclusion map] The inclusion map \begin{align*} I:C^{0,\gamma}(K;\mathbb R^m)\to C^{0,\alpha}(K;\mathbb R^m),\qquad I(h):=h \end{align*} is linear. It is also bounded: if $D=\operatorname{diam}(K)$, then for every $h\in C^{0,\gamma}(K;\mathbb R^m)$, \begin{align*} [h]_{C^{0,\alpha}(K)}\le \max\{1,D^{\gamma-\alpha}\}[h]_{C^{0,\gamma}(K)}. \end{align*} Thus \begin{align*} \|I(h)\|_{C^{0,\alpha}(K)}\le \max\{1,D^{\gamma-\alpha}\}\|h\|_{C^{0,\gamma}(K)}. \end{align*} The preceding steps show that every bounded sequence in $C^{0,\gamma}(K;\mathbb R^m)$ has a subsequence converging in $C^{0,\alpha}(K;\mathbb R^m)$. Therefore $I$ is a compact [linear map](/page/Linear%20Map), which is exactly the compact embedding \begin{align*} C^{0,\gamma}(K;\mathbb R^m)\hookrightarrow C^{0,\alpha}(K;\mathbb R^m). \end{align*} [/step]