Is refuting candidate deciders of the halting problem computable?
The usual proof that the halting problem is undecidable already gives you exactly such an algorithm. Given an algorithm $A$, we construct an algorithm $B$ that on input $x$ computes $A$ on program $x$ and input $x$, and then enters an infinite loop iff $A$ answered "halts". Now consider giving $B$ itself as an input. If $A(B,B)=\text{"halts"}$ then $B$ doesn't halt on $B$, so $A$ is wrong. Same if $A(B,B)=\text{"doesn't halt"}$. So $\langle B,B\rangle$ is an example input on which $A$ is wrong.