The metric completion of a normed space carries a unique compatible Banach-space structure

By [L2], the Cauchy classes carry termwise addition and scalar multiplication, and each class has a well-defined norm $\mathrm{\parallel }[x_n]\mathrm{\parallel }:={\lim }_n\mathrm{\parallel }{x}_n\mathrm{\parallel }$.

On constant sequences these operations agree with those of $X$, and $\mathrm{\parallel }i(x)\mathrm{\parallel }=\mathrm{\parallel }x\mathrm{\parallel }$ by definition, so $i$ is a linear isometry.

For classes $[x_n]$ and $[y_n]$, the distance of the published metric completion is ${\lim }_n\mathrm{\parallel }{x}_n-y_n\mathrm{\parallel }$, while the new difference class is $[x_n-y_n]$; therefore the completion metric is exactly the metric induced by the norm from step 1.1.

Because the underlying metric space is complete by [L1], step 2.2 makes $\widehat{X}$ complete for its new norm metric. So $\widehat{X}$ is Banach and $(\widehat{X},i)$ is a completion of $X$ in the sense of [L3].

Suppose a second Banach-space structure on the same underlying set induces the published completion metric and also makes $i$ a dense linear isometry. Its addition and scalar multiplication are continuous by [L4], and they agree with the operations of step 1.1 on the dense subset $i[X]\times i[X]$ and on $\mathbb{K}\times i[X]$; therefore they agree everywhere.

The norm is then forced as well, because in any compatible normed structure $\mathrm{\parallel }u\mathrm{\parallel }$ equals the metric distance from $u$ to $0$. So the compatible Banach-space structure is unique.