The statement seems intuitively true but is it? If so, how can I prove this?
If A is decidable and B is decidable, then A is Turing Reducible to B
computability;turing machines
One way of defining a Turing reduction (from $A$ to $B$) is as an algorithm that can decide $A$ if we have an oracle to decide $B$. The trick to notice here is that there is nothing that says you have to actually use the oracle.So given that $A$ is decidable, we can take the Turing machine that decides it, augment it with an oracle for $B$, but completely ignore the oracle and just run the TM as normal.