Show that $(x_{n})_{n}$ is dense in $[0,1]\setminus A$ then it is also dense in $[0,1]$

Let $B([0,1])$ be the borel set on $[0,1]$ and suppose that $\mu$ is a finite measure on $B([0,1])$. Then, define $A=[0,1]\cap \left \{ x;\mu(\left \{ x \right \})>0 \right \}$. I have shown that if $A$ is countable, then $[0,1]\setminus A$ is a separable metric space (because it is a subset of a separable metric space). Now if $(x_{n})_{n}$ is a dense sequence in $[0,1]\setminus A$, then I want to show that it is also dense in $[0,1]$. How can I prove it?

Solution 1:

If $I=(a,b)$ is an open interval in $[0,1]$ then $I \setminus A$ is uncountable so in particular a non-empty open subset of $[0,1]\setminus A$ and so contains some $x_n$ etc.