Note - this is not a duplicate, since this specific axiom has not been checked in any other answers.
Let $X$ be a metric space with inner product $(\cdot, \cdot)$ and let $\bar{X}$ be its completion. I want to define an inner product $(\cdot, \cdot)^{*}$ on the completion, such that $$ (x, y)^{*} = \lim_{n \to \infty} (x_{n}, y_{n}) $$ where for all $x \in \bar{X}$, I pick a unique sequence $x_{n}$ in $X$ that converges to it (in $\bar{X}$). I've been able to check all axioms other than $(x,x)^{*} = 0 \implies x = 0$. Can't $\lim_{n \to \infty} (x_{n}, x_{n})$ converge to $0$ without $x = \lim_{n \to \infty} x_{n}$ being $0$ (keeping in mind that it is possible that $\lim_{n \to \infty} x_{n}$ doesn't exist in the incomplete space $X$)
If $(x_n)$ is a null sequence then it is not possible that $((x_n),(x_n))^*\neq 0$ because $$ ((x_n),(x_n))^*=\lim_{n\to \infty }(x_n,x_n)=\lim_{n\to \infty }\|x_n\|^2=0\tag1 $$ By the other hand if $(x_n)$ is not a null sequence (that is, it doesn't converge to zero) then there exists some $\epsilon>0$ such that for any chosen $N\in \Bbb N$ there is some $j\ge N$ and $\|x_j\|\ge\epsilon$, so $\rm(1)$ cannot holds.