Many-one reducibility: why $A = \{1\}$ is not a counterexample to $\forall A : A \leq_m A \cup \{0\}$?
The property that, for all $x$, $x\in A$ iff $f(x)\in B$ means that $f(A)\subseteq B$ and $f(\overline{A})\subseteq \overline{B}$. Your proof attempt uses the condition $f(A)=B$, which is neither necessary nor sufficient.