If $X$ is compact metric then $B(X)$ is non-separabale I've seen the following argument:
Let $X$ be a compact metric space, then denote by $B(X)$ the set of bounded, measurable functions on $X$. $B(X)$ with the sup-norm is a $C^*$-algebra. Then, $B(X)$ is not separable.
Now, if I take $X=[0,1]$, and denote $A=span_{\mathbb{Q}}  \{\chi_{(a,b)} \forall a,b\in \mathbb{Q}\}$ then $A$ is a countable set in $B(X)$ and I had the (wrong) feeling that $A$ is dense in $B(X)$.
Because any measurable bounded function can be approximated uniformly by simple functions.I also didn't use the compactness of $X$. 
So, why am I wrong and I'll be happy to see a proof for non-separability of $B(X)$.
 Thank you in advance.

If  $X$  is compact metric space  such that there exists a countable partition $(X_k)_{K \in N}$ of $X$  into non-emtly measurable subsets of $X$. Then $B(X)$ is non-separabale  w.r.t. sup-norm.  Indeed consider the family of measurable functions $(f_J)_{J \subseteq N}$  where $f_J$ is an indicator  function of $\cup_{k \in J}X_k$ for each $J \subseteq N$. then $\sup_{x \in X}|f_{J_1}(x)-f_{J_2}(x)|=1$ for each different $J_1,J_2 \subseteq N$. This means that $B(X)$ is non-separable.    
