Let $f(x, y) = x^m y^n - x^n y^m$. If $n = m$, the problem statement will be fulfilled no matter how we choose $B$ so from now on, without loss of generality, we consider $n > m$.
Since $m$ and $n$ are both odd, we have that $n - m$ is even and we get
$$\begin{align*} f(x, y) &= x^m y^m (y^{n-m} - x^{n-m}) \\ f(x, y) &= x^m y^m (y^2 - x^2) Q(x, y) \\ f(x, y) &= x^m y^m (y - x)(y + x) Q(x, y) \end{align*}$$
where $Q(x, y) = y^{n-m-2} + y^{n-m-4} x + \dots + y^{n-m-2}$.

Now if one of $x, y$ is even, $f(x, y)$ is even. If both are odd, then $f(x, y)$ is again even since $x + y$ and $x - y$ are even in that case. This shows that we only need to consider divisibility by $5$.

If $A$ contains at least one element divisible by $5$, we can put it in $B$ and that will give us the solution easily.

Now we consider the case when none of the elements in $A$ is divisible by $5$. If some two numbers in $A$ give the same remainder modulo $5$, we can choose them and then $x - y$ will be divisible by $5$ which solves the problem.

Now we consider the case when all remainders modulo $5$ in $A$ are different. Take a look at the pairs $(1, 4)$ and $(2, 3)$.
Since we have three different remainders modulo $5$, by pigeonhole principle one of these pairs has to be completely in $A$ (when elements are considered modulo $5$). Then if we pick the numbers from $A$ that correspond to those two remainders we get that $x + y$ is divisible by $5$ so the problem statement is fulfilled again. This completes the proof.