Is $BC([0,1))$ ( space of bounded real valued continuous functions) separable? Is $BC([0,1))$ a subset of $BC([0,\infty))$? It is easy to prove the non-separability of BC([0,$\infty$)) and the separability of C([0,1]). It seems to me we can argue from the fact that any bounded continuous function of BC([0,$\infty$)) must also be in BC([0,1)) to somehow show BC([0,1)) is not separable, but  BC([0,1) 
 A: The space $BC([0,1))$ is not separable. For the begining consider function
$$
\varphi(x)=\max(1-|2x|,0)
$$
Then for each binary sequence $s\in\{0,1\}^\mathbb{N}$ we define function
$$
g_s(x)=\sum\limits_{k=1}^\infty s(k)\varphi\left(x-\frac{k}{2}\right)
$$
This is uncountable family in $BC([0,+\infty))$. Moreover if $s'\neq s''$, then $\Vert g_{s'}-g_{s''}\Vert_\infty=1$. Now consider functions
$$
f_s(x)=g_s\left(\frac{1}{1-x}\right), \quad s\in\{0,1\}^\mathbb{N}
$$
It is easy to see that $\{f_s:s\in\{0,1\}^\mathbb{N}\}$ is uncountable subset of $BC([0,1))$ and if $s'\neq s''$, then $\Vert f_{s'}-f_{s''}\Vert_\infty=1$. This is impossible if $BC([0,1))$ separable, so $BC([0,1))$ is not separable.
A: $BC([0,1))$ is not a subset of $BC([0,\infty))$; in fact, these two sets of functions are disjoint. No function whose domain is $[0,1)$ has $[0,\infty)$ as its domain, and no function whose domain is $[0,\infty)$ has $[0,1)$ as its domains. What is true is that $$\{f\upharpoonright[0,1):f\in BC([0,\infty))\}\subseteq BC([0,1))\;.$$
There is, however, a very close relationship between $BC([0,\infty))$ and $BC([0,1))$, owing to the fact that $[0,\infty)$ and $[0,1)$ are homeomorphic. An explicit homeomorphism is $$h:[0,\infty)\to[0,1):x\mapsto \frac2\pi\arctan x\;.$$ This implies that $BC([0,1))$ and $BC([0,\infty))$ are actually homeomorphic, via the map $$H:BC([0,1))\to BC([0,\infty)):f\mapsto f\circ h\;,$$ as is quite easily checked. Thus, one of $BC([0,\infty))$ and $BC([0,1))$ is separable iff the other is.
A: Consider the following, much simpler, construction:
For each binary sequence $a \in \{0,1\}^{\mathbb{N}}$ define bounded and continuous function $f_{a}(x)$ such that, $$f_{a}\left(\frac{1}{k}\right) = \begin{cases}1, \hspace{0.2cm} \text{if} \hspace{0.2cm} a_k = 1, \\
 0, \hspace{0.2cm} \text{if}\hspace{0.2cm} a_k=0 \end{cases}.$$
In the other points of $(0,1)$ define $f$ to be linear function. More precisely in the intervals $\left(1-\frac{1}{k}, 1-\frac{1}{k+1}\right)$ define $f$ to be the linear function through the points $$\left(\frac{1}{k},f_a \left(\frac{1}{k}\right)\right),\left(\frac{1}{k+1},f_a \left(\frac{1}{k+1}\right)\right).$$
Now this saw-shaped functions are obviously continuous and bounded in the open interval $(0,1)$.
Thus we have constructed a uncountable family $\{f_{a}(x)\}_{a \in \{0,1\}^{\mathbb N}} \subset \operatorname{BC}(0,1).$
$a \ne b \in \{0,1\}^\mathbb N$ we get $f_a(x),f_b(x) \in \operatorname{BC}(0,1),$ for which it is trivial that
$$\|f_a (x) - f_b(x) \|_\infty =\sup\{|f_a(x)-f_b(x)| : x \in (0,1)\}=1.$$
And so for examples take the balls in $\operatorname{BC}(0,1)$ centered at these functions with radius $1/2$. $\left\lbrace B_{\frac{1}{2}}(f_{a})\right\rbrace_{a \in \{0,1\}^{\mathbb N}} \subset \operatorname{BC}(0,1).$ This is uncountable family and it is not separable. Thus the whole space $\operatorname{BC}(0,1)$ cannot be separable! (Or else we'd get a injective map between countable and uncountable sets, which is absurd!)
I'd like to point out that this is isometric embedding of , so to speak the minimal not separable part of $\ell_\infty, i.e. \mathcal{F} : \{0,1\}^{\mathbb{N}} \hookrightarrow \operatorname{BC}(0,1), \hspace{0.2cm} \mathcal{F}(a):=f_a.$

