(1) Assume $D$ compact then by Weiertrass approximation you get that polynomials are dense in $C(D)$, since $D\subset B_R(0)$ for some $R>0$ and polynomials are $\alpha$-Hölder in $B_R(0)$ we get the result.
If $D$ is not compact the result is not true: Take $D=(0,1)$ and $f(x)=\sin(1/x)$. Take $0 and $g\in C^{0,\alpha}$ such that $\| g-f\|_\infty , then for every $n\in \mathbb{N}$ there exists $x_n,y_n\in (0,1)$ with $|x_n-y_n| and $f(x_n)=1$, $f(y_n)=-1$. Then $|g(x_n)-g(y_n)| \geq 2(1-r)>r^{\alpha}>|x_n-y_n|^{\alpha}$ if $r$ is small enough, contradicting $g\in C^{0,\alpha}$.
If $D=\mathbb{R}$ a similar argument works if we use something like $f(x)= \sin(2n\pi(x-n))$ if $x\in [n,n+1/n]$, $n\in \mathbb{N}$ and zero otherwise.
(2) We know that $C(K)$ is complete so we only need to check that the limit function of a Cauchy sequence in $C^{0,\alpha}$ is in this space and that the seminorms converge. For this notice $ |f(x)-f(y)|\leq |f(x)-f_n(x)|+|f_n(x)-f_n(y)|+|f_n(y)-f(y)| $ So, picking $n$ so that $\| f-f_n\|_{\infty} <\epsilon$ we get, and remembering that $\frac{|f_n(z)-f_n(w)|}{|z-w|^\alpha} \leq [f_n]_\alpha \leq C$, $ \frac{|f(x)-f(y)|}{|x-y|^\alpha} \leq C+2\epsilon/|x-y|^\alpha $ which gives $[f]_\alpha .
To see that the seminorms converge just notice $ \frac{|f(x)-f_n(x)-f(y)+f_n(y)|}{|x-y|^\alpha} \leq\limsup_m [f_m-f_n]_\alpha $ which gives $ \limsup_n [f-f_n]_\alpha \leq \limsup_n \limsup_m [f_m-f_n]_\alpha \to 0. $ (3) I don't know about this statement as is, but Arzela-Ascoli gives that bounded closed sets in $C^{0,\alpha}(K)$ are precompact in $C(K)$