How to see that $\mathcal C_{(0)}[0,\infty)$ is separable? 
Define the subspace $\mathcal C_{(0)}[0,\infty)\subset (\mathbb
 R^d)^{[0,\infty)}$ as follows
$$\mathcal C_{(0)}[0,\infty):=\{w:[0,\infty)\to \mathbb R^d: w \text{
 is continuous},w(0)=0\}$$
If we equip $\mathcal C_{(0)}$ with the locally uniform convergence 
  it becomes a complete separable metric space.

If we consider for instance $\mathcal C_{(0)}[0,1]$ then it follows from Weierstrass approximation theorem that there's a dense, countable subset. 
In this case, since my functions are not defined on a compact interval, how can I see that the space is indeed separable? Is there some extension of Weierstrass theorem for non-compact intervals?
Thanks in advance.
 A: Let us consider the bigger space $\mathcal C([0,\infty))$ and show that it is separable. As Paul Sinclair remarked in his comment, it suffices to consider $d = 1$.
We know that each $\mathcal C([0,n])$, $n \in \mathbb N$, contains a countable dense set $D_n$. Define 
$$j_n : \mathcal C([0,n]) \to \mathcal C([0,\infty)), j_n(\phi)(t) = \begin{cases} \phi(t) & t \le n \\ (n+1-t)\phi(n)  & n \le t \le n+1 \\ 0 & t \ge n+1 \end{cases}$$
We show that $D = \bigcup_{n=1}^\infty j_n(D_n)$ is dense in $\mathcal C([0,\infty))$.
Given $f \in \mathcal C([0,\infty))$, choose $d_n \in D_n$ such that $\lVert f\mid_{[0,n]} - d_n \rVert_\infty < 1/n$. We claim that $d'_n = j_n(f_n) \to f$ in the topology of locally uniform convergence. So let $t_0 \in [0,\infty)$. Let $n_0 \in \mathbb N$ such that $t_0 < n_0$. Then $U = [0,n_0)$ is an open neigbhorhood of $t_0$. For $n  \ge n_0$ we have $\sup\{ \lvert f(t) - d'_n(t) \rvert \mid t \in U\} \le \sup\{ \lvert f(t) - d'_n(t) \rvert \mid t \in [0,n]\} = \sup\{ \lvert f(t) - d_n(t) \rvert \mid t \in [0,n]\} = \lVert f\mid_{[0,n]} - d_n \rVert_\infty < 1/n$ which proves our claim.
