How to prove the language of all Turing Machines that accept an undecidable language is undecidable?
I think you are on the right track but things need to be made more explicit. First, what machine exactly is $A$? Any machine? Where does your contradiction appear?
First, you need to explicitly say that $A$ is a machine for an undecidable language, for example, let $A$ be the universal machine, that on inputs $\langle M \rangle w$ simulates $M$ on $w$, and accepts if $M$ accepts $w$.
Then, say explicitly that $S$ is a decider (a machine that always halts), and include the simple extra steps that show that $S$ accepts $\langle M \rangle w$ if and only if $w \in L(M)$. Finally, recall that $S$ is then a decider for a problem we already know is undecidable; a contradiction.
Edit: Sorry, it seems that $A_{TM}$ is standard notation for the universal machine. Forget about that comment if that is the notation used in your context.