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) 

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 $\ell_\infty, i.e. \mathcal{F} : \{0,1\}^{\mathbb{N}} \hookrightarrow \operatorname{BC}(0,1), \hspace{0.2cm} \mathcal{F}(a):=f_a.$

