Show that a decidable language is not decided by a decider in a given set
Suppose that $A$ is a recognizable set containing descriptions of always halting Turing machines. We want to show that there exists a decidable language $L$ such that for any $M$ deciding $L$ we have $\langle M\rangle\notin A$.

We can construct such $L$ via simple diagonalization. Assume $A$ is infinite (otherwise the problem is trivial). We will define a Turing machine $M$ such that $L(M)$ does not have a decider in $A$. Let $M_A$ be an enumerator of $A$. Given input $x$, $M$ executes $M_A$ untill it outputs the $|x|'th$ word in the enumeration $\langle M_{|x|}\rangle$. Now, execute $M_{|x|}$ on $x$ and flip its answer.

I leave it to you to show that $M$ always halts and that $L(M)$ disagrees with $L(M_i)$ on at least $2^i$ inputs, where $M_i$ is the i'th Turing machine in the enumeration of $A$.