Dense subspace of Zygmund space or Hölder space? Do we know any function spaces dense in  Zygmund space $C_*^s$(a special case of Besov space, i.e. $C_*^s = B^s_{\infty,\infty}$) or Hölder space$C^{k,r}$, with underlying field $\mathbb{R}^d$?
Will $C^\infty$ do the job?
 A: If a space is not separable, you should not expect it to have any nice dense subspaces, such as $C^\infty$. Indeed, $C^\infty$ is separable with respect to any Zygmund- or Hölder-norm; a countable dense subset can be constructed from a countable partition of unity multiplied by polynomials with rational coefficients. 
For example, there is no (norm) dense subspace of $L^\infty$ that is worth mentioning. 
None of the spaces you mention are separable. For example, the functions $\max(x-s,0)^{k+\alpha}$ for $s\in\mathbb R$ form an uncountable set in which any two points are at distance at least $1$ in the $C^{k,\alpha}(\mathbb R)$ norm. Same example (with $\alpha=1$) works for Zygmund spaces.
For $0<\alpha<1$, the closure of $C^\infty$ functions in the Hölder space $C^{k,\alpha}$ is  known as the  little Hölder space $c^{k,\alpha}$. Its elements are described by the $k$th derivative having the modulus of continuity $\omega(\delta)$ such that $$\lim_{\delta\to 0}\frac{\omega(\delta)}{\delta^\alpha}= 0 \tag1$$ Indeed, (1) holds for $C^\infty$ functions and is preserved under convergence in the norm. 
When $\alpha=1$, the closure of $C^\infty$ functions in $C^{k,1}$ is simply $C^{k+1}$. The reason is that convergence of $C^\infty$ functions in $C^{k,1}$ norm implies uniform convergence of their derivatives of order $k+1$. Thus, the continuity of $(k+1)$th derivative is preserved. 
The situation in the Zygmund spaces is similar to what happens what $\alpha$ is fractional: the closure of $C^\infty$ is the space of functions whose second modulus of continuity $\omega_2$ satisfies a vanishing condition instead of the boundedness condition. This is also called a little Zygmund space. For example, functions in $c^1_*$ satisfy 
$$\lim_{\delta\to 0}\sup_x \sup_{|h|<\delta}\left| \frac{f(x+h)-2f(x)+f(x-h)}{h} \right|=0 \tag2$$
Zygmund himself observed the difference between the big and little spaces on the first page of his paper Smooth functions which you can read without subscription. 
