We will first prove that $|A \cup B| \le 66$, or equivalently $|A| \le 33$. For this purpose, we only need to prove that, if $A$ is a subset of $\{1, 2, \dots, 49\}$ with 34 elements, then there must exist $n \in A$ such that $2n + 2 \in A$. The proof is as follows.

Divide $\{1, 2, \dots, 49\}$ into 33 subsets:

$\{1, 4\}$, $\{3, 8\}$, $\{5, 12\}$, $\dots$, $\{23, 48\}$, 12 subsets;
$\{2, 6\}$, $\{10, 22\}$, $\{14, 30\}$, $\{18, 38\}$, 4 subsets;
$\{25\}$, $\{27\}$, $\{29\}$, $\dots$, $\{49\}$, 13 subsets;
$\{26\}$, $\{34\}$, $\{42\}$, $\{46\}$, 4 subsets.

By the Pigeonhole Principle we know that there exists at least one subset with 2 elements among them which is also a subset of $A$. That means there exists $n \in A$ such that $2n + 2 \in A$.

On the other hand, let

$A = \{1, 3, 5, \dots, 23, 2, 10, 14, 18, 25, 27, 29, \dots, 49, 26, 34, 42, 46\}$

$B = \{2n + 2 \mid n \in A\}$.

We find that $A$ and $B$ satisfy the condition and $|A \cup B| = 66$.

Answer: B.