On the necessity of being a *dense* subset in completion of a metric space Quoted from the book Introductory functional Analysis by Erwin Kreyszig : 

1.6-2 Theorem (Completion). For a metric space $X = (X, d)$ there 
  exists a complete metric space $\bar{X}=(\bar{X}, \bar{d})$ which has a subspace $W$ that is 
  isometric with $X$ and is dense in $\bar{X}$. This space $\bar{X}$ is unique except for 
  isometrics, that is, if $\bar{X'}$ is any complete metric space having a dense 
  subspace $W'$ isometric with $X$, then $\bar{X'}$ and $\bar{X}$ are isometric.

Why $W$ must be a dense subset of $\bar{X}$? Is it just because generalization is from completion of $\mathbb{Q}$ (to $\mathbb{R}$) or being dense is something to do with a super-space to be complete? And if so, why? 
Edit - For example (0,1) is not a complete space but its closure [0,1] is. And still (0,1) is not dense in [0,1] because int((0,1)) is not empty set.     

I am not sure what is your question. 
Why $W$ must be dense? 
This is by the construction in the proof of the theorem. 
Why is it important? 
Because it basically tells you that in every metric space, only adding a 'few' limit points will make it complete. This is remarkable! Assume for a second you ignore the density part. Then a good question would be: given a metric space, what is the smallest complete space containing it? The answer we now know that it is it's 'closure'.  
