$L_\infty[0,1]$ completness and separability 
Prove that $L_\infty[0,1]$ with the norm given by
$$\Vert f\Vert_\infty:= \inf\{S(N): \mu(N)=0\}, \quad \mbox{where}
 \quad S(N)=\sup\{|f(x)|: x\notin N\}.$$
is complete and is not separable.

I wasn't able to find an uncountable family of functions $(f_i:[0,1] \longrightarrow \overline{\mathbb{R}})$ such that $\Vert f_i \Vert_\infty= 1$ to prove that $L_\infty[0,1]$ is not separable.
And I need a hint to prove the completness. Can anybody help me?
 A: $L_\infty[0,1]$ is not separable
PROOF  Assume that $L_\infty[0, 1]$ 
is separable. Now consider the set 
$$
F= \{ f_t : f_t(x)= \chi_{(0,t)} (x)  \ \ t \in (0,1) \} 
$$
Since all the $f_t$ are measurable and clearly $\|f_t\|_\infty=1<\infty$, then $F\subset L_\infty[0,1]$. Of course since $(0,1)$ is uncountable,  $F$ is also uncountable. 
Since $L_\infty[0, 1]$ is separable there exist a countable subset $A=\{ a_1, a_2, \cdots \}\subset L_\infty[0,1]$ such that $\overline{A} = L_\infty[0,1]$. Take any  $f_t \in F \subset \overline{A}$, and for $\varepsilon=1/2$ choose $a \in A$ such that $\| f_t-a\|_\infty < 1/2$. Then for $t' \neq t$
$$
1=\|f_t-f_t' \|_\infty \leq \|f_t-a\|_\infty + \|a-f_t' \|_\infty < 1/2 + \|a-f_t' \|_\infty
$$
which gives that $ 1/2 + \|a-f_t'\|_\infty>1  \Longrightarrow \|a-f_t' \|_\infty>1/2$. Thus for every $f_t \in F$ there exist only and only one $a \in A$ such $\| f_t-a\|_\infty < 1/2$, that is $F$ and a subset of $A$ have the same elements, but since $F$ is uncountable, $A$ must be uncountable, a contradiction. $\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ {\blacksquare}$
