Prove $\{xy: x \in A \land y \in B \land |x| = |y|\}$ is context-free
There are many ways to prove this. Here is another one, which proves a stronger result: $A \diamond B$ is linear.

Suppose that both $A,B$ are languages over a common alphabet $\Sigma$. Let $\langle Q_A,q_{0A},\delta_A,F_A \rangle$ and $\langle Q_B,q_{0B},\delta_B,F_B \rangle$ be DFAs for $A^R$ (the reverse of $A$) and for $B$. We construct a linear grammar with a nonterminals $\{S\} \cup Q_A \times Q_B$, with the following rules:


For every $f_A \in Q_A$ and $f_B \in Q_B$, the rule $S \to \langle f_A, f_B \rangle$.
For every $q_A \in Q_A$, $q_B \in Q_B$, and $\sigma_A,\sigma_B \in \Sigma$, the rule $\langle \delta_A(q_A,\sigma_A), \delta_B(q_B,\sigma_B) \rangle \to \sigma_A \langle q_A, q_B \rangle \sigma_B$.
The rule $\langle q_{0A},q_{0B} \rangle \to \epsilon$.


In fact, your proof also shows that $A \diamond B$ is linear, since $\{a^n \# b^n : n \geq 0 \}$ is easily seen to be linear, and all the closure properties you use also hold for linear languages.