The completion of a separable inner product space is a separable Hilbert space

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Let $X$ be a separable inner product space and $\bar X$ its completion. Then I want to check that $\bar X$ is a separable Hilbert space.

But I am stuck here, because how do you prove that $\bar X$ has a inner product induced by the one of $X$?, and how do you construct a complete orthonormal system for $\bar X$? because the completion is a set of representatives, that we don't even know if they look as the elements of $X$.

But I am a little bit confused and stuck,I think more hypothesis on $X$ to get the result, because well, we know that $\bar X$ by definition, but how do you ensure that it has a inner product? and even more, how do you construct a countable dense set from elements that are abstract?, since we have to recall that all the points that live in the completion are equivalency classes of sequences, so it is hard for me to deal with them.

I know that the completion is the quotient space of the Cauchy sequences of $X$ and the ones that converge to $0$ and $\bar X$ is complete with respect to the metric defined as

$$\lim_{n \to \infty} d(x_n,y_n)$$

but I don't know how to get the result with that.

So Can someone help me to prove this assertion please?