I have to prove 0.v=0 where the left 0 is a zero vector and v is a vector and the right 0 is a number. Is the following attempt right given that I am just allowed to use the axioms of inner product?
Left hand side: 0.v= (u+(-u)).v where u is a vector and -u is its additive inverse. Then we get, u.v+(-u).v=u.v-u.v=0 Hence proved.
Your proof is perfectly right. You can also have used the left-linearity of the dot product: $$0\cdot u=0\times (0\cdot u)=0.$$ This used the fact that $0$ is absorbing for $\times$, which relies on the same trick you used: $$0\times x=(1-1)\times x=x-x=0.$$