Prove that there is no computable enumeration of all decidable languages
WLOG assume we are talking about languages over a binary alphabet $\{0,1\}$. Let $M'$ be an enumerating Turing Machine for $L$, i.e., every element of $L$ is eventually output by $M'$. Let $\langle M_j \rangle$ denote the $j$th TM output by $M'$. Define
$$L_D = \{ x \mid M_x(x) \text{ does not accept} \},$$
where $x$ is interpreted as a number in base 2. Note that $L_D$ is decidable: given $x$ you can decide if $x \in L_D$ by running $M'$ until it outputs $x$th TM $M_x$ and then running $M_x$ on $x$ and flipping the output. This is a decider since all TMs output by $M'$ halt on all inputs. Thus $L_D$ appears in the enumeration of R, say, as $L_k$ for some $k$.

It is left to show that $L \cap A_k = \emptyset$. Suppose that there exists a TM $M$ such that $\langle M \rangle \in L \cap A_k$. Then let $\ell$ be the smallest number such that $\langle M \rangle$ is output as the $\ell$th element by the enumerator $M'$, i.e. $M = M_\ell$. Therefore, we have $\ell \in L(M) \iff \ell \in L(M_\ell) \iff \ell \notin L_D \iff \ell \notin L_k \iff \ell \notin L(M)$, which is a contradition.