Suppose we are given a Turing machine $e$ terminates with result $0$ or $1$ on all inputs, such that $e(n) = 0$ when $n \in A$ and $e(n) = 1$ when $n \in B$. Consider the algorithm $F$ that returns a Turing machine for a given natural number:$halt(k)$ is the machine that terminates with result $k$ on every input.

Because $e$ terminates on all inputs, so does $F$. By construction $e(F(n)) \ne e(n)$ for each input $n$: Consider the case that $e(n) = 0$, then $F(n) = halt(1)$. $halt(1) \in B$ because

but then $e(F(n)) = e(halt(1)) = 1 \ne 0 = e(n)$. The same argument applies to $e(n) = 1$.

By the second recursion theoremhttps://www.math.ucla.edu/~ynm/lectures/2009csl.pdf, there exists a Turing machine $f$ realizing $F$ applied to its own Gödel number, that is, $\phi_{\ulcorner f \urcorner} \simeq \phi_{\ulcorner F(f) \urcorner}$. Because $F(f)$ is $halt(0)$ or $halt(1)$, both constant, $\phi_{\ulcorner f \urcorner}$ is constant as well, say with value $k$. Hence

so either both $f$ and $F(f)$ lie in $A$, or both lie in $B$. In either case $e(f) = e(F(f))$. However, $e(n) \ne e(F(n))$ for all $n$.