Dense subset of given space If $E$ is a Banach space, $A$ is a subset such that 
$$A^{\perp}:= \{T \in  E^{\ast}: T(A)=0\}=0,$$ then $$\overline{A} = E.$$
I don't why this is true. Does $E$ has to be Banach? Thanks

Did you mean to say that $A$ is a vector subspace, or does $\overline{A}$ mean the closed subspace of $E$ generated by $A$?  If $A$ were only assumed to be a subset and $\overline{A}$ means the closure, then this is false.  E.g., let $A$ be the unit ball of $E$.  
Suppose that the closed subspace of $E$ generated by $A$, $\overline{\mathrm{span}}(A)$,  is not $E$.  Let $x\in E\setminus \overline{\mathrm{span}}(A)$.  Using Hahn-Banach you can show that there is an element $T$ of $E^*$ such that $T(A)=\{0\}$ and $T(x)=1$.  (Start by defining $T$ on the subspace $\overline{\mathrm{span}}(A)+\mathbb C x$.)  No, this does not depend on $E$ being complete.
