Show that a set of $\alpha$-Holder functions is compact 
Let $(K,d_K)$ be a compact metric space and consider $Z=\mathbb{R}^d$, with the usual metric. Let $L,\alpha>0$ and $(t,z_0) \in K\times \mathbb{R}^d$. Show that the set $$\mathcal{H}_{L,\alpha,t_0,z_0}=\{f:K\rightarrow\mathbb{R}^d:\ f \text{ is } \alpha\text{-Holder with constant } L \text{ and } f(t_0)=z_0\}.$$ is a compact set of $C:=C((K,Z),d_C)$.

It is clear that $\mathcal{H} \subset C$. Showing compactness, for me, is the hard part. First of all, I tried to show that $\mathcal{H}$ is bounded and closed, then, I realized that I do not have that $\mathcal{H} \subset\mathbb{R}^d$, so I tried other path: showing that $\mathcal{H}$ is sequentially compact. My proof comes and I am not certaing of it, was:


Consider $\{f_n\} \subset \mathcal{H}$ a sequence. Then, for $x \in K$, $$||f_n(x)-z_0||\leq L\cdot d_K(x,z_0)^\alpha$$ Since $K$ is compact, then, bounded, the sequence $\{f_n(x)-z_0\} \subset \mathbb{R}^d$ is bounded and so it is $\{f_n(x)\} \subset \mathbb{R}^d$, since $z_0$ is fixed. Then, it has a convergent subsequence $\{f_{n_k}(x)\}$, where $f_{n_k}(x) \rightarrow f(x)$ for all $x \in K$.From here, I concluded that $f_{n_k} \rightarrow f$, with no more arguments (and I do not think is such direct). Then, the problem was just showing that $f \in \mathcal{H}$, which is also clear.


Could someone give me a hand with other solution or showing me what am I missing?
 A: I assume that $d_C$ is the $\sup$-metric. Let $f \in \mathcal{H}$ be arbitrary. Then, for all $x \in K$
$$
\lVert f(x) \rVert \leq \lVert f(t_0) \rVert + \lVert f(t_0) - f(x) \rVert \leq \lVert z_0 \rVert + Ld(t_0, x)^\alpha \leq \lVert z_0 \rVert + LC^\alpha,
$$
where $C := \sup_{y, z \in K} d(y, z)$. $C$ is finite, because $K$ is compact and therefore bounded. So all coordinate functions $f_i$, $i \in \lbrace 1, ..., d \rbrace$ are bounded by the same constant $\lVert z_0 \rVert + LC^\alpha$ that is independent of $f$.
Now let $\varepsilon > 0$ and $\delta := \left(\frac{\varepsilon}{L}\right)^\frac{1}{\alpha}$. Then, we have for all $x, y \in K$ that fulfil $d(x, y) < \delta$:
$$
\lvert f_i(x) - f_i(y) \rvert \leq \lVert f(x) - f(y) \rVert \leq Ld(x, y)^\alpha  < L\delta^\alpha = \varepsilon
$$
So all coordinate functions are uniformly continuous with $\delta$ chosen independently  of $f$.
So according to Arzelà-Ascoli's theorem, for every sequence $(g^{(j)})_{j \in \mathbb{N}} \subseteq \mathcal{H}$, there exists a subsequence (that we again denote by $g^{(j)}$) and some $g \in C(K, Z)$ such that the former subsequence converges to $g$ in $d_C$. We have to prove that $g \in \mathcal{H}$. But this is easy: Convergence in $d_C$ implies pointwise convergence and therefore for all $x, y \in K$ where $x \neq y$:
$$
\frac{\lVert g^{(j)}(x) - g^{(j)}(y) \rVert}{d(x, y)^\alpha} \leq L \overset{j \rightarrow \infty}{\implies} \frac{\lVert g(x) - g(y) \rVert}{d(x, y)^\alpha} \leq L
$$
This is Hölder-continuity. Especially we have
$$
g^{(j)}(t_0) = z_0 \overset{j \rightarrow \infty}{\implies} g(t_0) = z_0.
$$
We are done.
Remark: I know that Wikipedia only states Arzelà-Ascoli's theorem for $d = 1$. But you can construct a convergent subsequence for the first coordinate function, then a sub-subsequence for the second and so on.
