Your whole methodology is flawed, so far as I can tell. The generally accepted way to show that the zero vector is unique, is to assume there's another vector that satisfies the behavior of the zero vector, and that, therefore, it must equal the usual zero vector. That is, using your notation, you assume both that there is a vector such that for all and that the usual zero vector satisfies for all
Now you must SHOW that In your proof, at one point, you assumed and went on to show a contradiction! That logic doesn't work at all. This is not a proof by contradiction. Use the two equations you have assumed to show the required equation is true.