It is a known theorem that if B is a recursive set and A many-to-one reduces to B, then A is also a recursive set.
I am looking for a counterexample for the converse, so a non-recursive set B such that a recursive set A many-to-one reduces to it.
I have already proved that B must be recursively enumerable.
Let $B$ be any non-empty proper subset of $\mathbb N$ and let $p \in B, q \in \mathbb N \setminus B$ be arbitrary but fixed. Let $A$ be a recursive subset of $\mathbb N$ and define
$$f(x) = \begin{cases} p &\text{if }\ x \in A \\ q &\text{otherwise} \end{cases}$$
Then $f$ is recursive and $x \in A \iff f(x) \in B$, i.e. $A$ is many-to-one reducible to $B$.