Independent set problem
The reason you can't just take $G$ as an input to $A$ is that (strangely enough) $A$ only works on connected graphs.

You want to prove that if $A$ exists, then Independent Set is solvable in polynomial time.  The reason you can prove that is by using a polynomial time reduction.   You have proved that if $A$ exists, then $A$ can solve Independent Set on any graph by using your trick of adding a universal vertex.

You have also proved that Independent Set on non-connected graphs is not particularly much easier than on connected graphs.