Let $Y$ and $Z$ be completions of $X$, then there exists an isometry from $Y$ onto $Z$.

Question

Proof Attempt: Let $X$ be dense on metric spaces $Y$ and $Z$. Let $\overline{X}=Y$ and $\overline{X}=Z$. Suppose that $\{a_n\}$ is a sequence in $A$ such that $\{a_n\}\rightarrow y,z$ such that $y\in Y$ and $z\in Z$. Let $M\in\Bbb{N}$ be such that $\forall i\geq M$, we have $d_Y(y,a_i),d_Z(z,a_i)\lt\epsilon/2$. Then, we extend the identity mapping on $X$, by defining $i(y)=z$. Hence, we have \begin{align} \vert d_Y(y,a_i)-d_Z(i(y),i(a_i))\vert=\vert d_Y(y,a_i)-d_Z(z,a_i)\vert\lt\epsilon \end{align} This means $\forall y\in Y$, $\forall z\in Z$, and for all sequences $\{a_n\}\rightarrow y,z$ in $X$, we have $d_Y(y,a_i)=d_Z(z,a_i)$ - an isometry.