If you allow $A$ to contain an interval, let $$A = (\mathbb Q\cap [0,1])\cup [0,1/2].$$ If not, consider $q_i$ the $i$th rational number in some enumeration of $\mathbb Q\cap [0,1]$, and let $$A=\bigcup^\infty_{i=1} (q_i-4^{-i}, q_i+4^{-i})$$

If you allow $A$ to contain an interval, let $$A = (\mathbb Q\cap [0,1])\cup [0,1/2].$$ If not, consider $q_i$ the $i$th rational number in some enumeration of $\mathbb Q\cap [0,1]$, and let $$A=\bigcup^\infty_{i=1} (q_i-4^{-i}, q_i+4^{-i})\cap (\mathbb R\setminus\mathbb Q)$$

Sure, let $$A = (\mathbb Q\cap [0,1])\cup [0,1/2].$$

If you allow $A$ to contain an interval, let $$A = (\mathbb Q\cap [0,1])\cup [0,1/2].$$ If not, consider $q_i$ the $i$th rational number in some enumeration of $\mathbb Q\cap [0,1]$, and let $$A=\bigcup^\infty_{i=1} (q_i-4^{-i}, q_i+4^{-i})$$