[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*}
[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]
[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]
