Verify these claims. For the first consider ∣0⟩+∣0′⟩ and use the advertised properties of the two null vectors in turn. For the second start with ∣0⟩=(0+1)∣V⟩+∣−V⟩. For the third, begin with ∣V⟩+(−∣V⟩)=0∣V⟩=∣0⟩. For the last, let ∣W⟩ also satisfy ∣V⟩+∣W⟩=∣0⟩. Since ∣0⟩ is unique, this means ∣V⟩+∣W⟩=∣V⟩+∣−V⟩. Take it from here.