Why is Completeness and Compactness not equivalent in Normed Spaces? Given a complete normed space $X=(X,\|\cdot\|)$.  Every Cauchy sequence converges in it. I am not able to understand why we can't show that every bounded sequence in $X$ will have a convergent subsequence.
Please give an example to clarify why completeness does not imply compactness and do explain where does the problem lie.

A normed space $X=(X,\|\cdot\|)$ is a metric space with the metric induced by the norm
\begin{align*}
d(x,y)=\|x-y\|
\end{align*}
All concepts defined for metric spaces are applicable, in particular, to normed spaces. Therefore we can also use the results of completeness and compactness derived in metric spaces.
The following facts stated without proof might help to better see the relationship of completeness and compactness in metric spaces and therefore also in normed spaces.

The following is valid:
  
  
*
  
*A metric space $(X,d)$ is complete if every Cauchy sequence in $X$ has a convergent subsequence.
  
*In a metric space $(X,d)$ compactness and sequential compactness are equivalent.
We see that in a metric space $(X,d)$ compactness is a stronger property than completeness, since we have
\begin{align*}
\text{ compact } \quad\Longleftrightarrow\quad \text{ sequentially compact } \quad\Longrightarrow\quad \text{ complete }
\end{align*}

The  implication in the other direction does not hold, since e.g. $\mathbb{R}$ is complete in the usual euclidean metric, but not compact. 
So, it is reasonable to ask what extra condition we could impose to a complete metric space in order to guarantee its compactness. Such a condition is total boundedness.

A metric space $(X,d)$ is said to be totally bounded if for every $\varepsilon>0$, there is a finite covering of $X$ by $\varepsilon$-balls.
  
  
*
  
*Example: Under the metric $d(x,y)=|x-y|$, the real line $\mathbb{R}$ is complete but not totally bounded, while the subspace $(-1,1)$ is totally bounded but not complete. The subspace $[-1,1]$ is both complete and totally bounded.
  
  
  Theorem: A metric space $(X,d)$ is compact if and only if it is complete and totally bounded.
\begin{align*}
\text{ compact } \quad\Longleftrightarrow\quad \text{ complete } +\text{ totally bounded }
\end{align*}
If $X$ is a compact metric space, then  $X$ is complete as noted above. The fact that $X$ is totally bounded is a consequence of the fact that the covering of $X$ by all open $\varepsilon$-balls must contain a finite subcovering.
The other direction can be shown, by proving that completeness and total boundedness implies sequential compactness, from which compactness follows.

Note: The answer above is based upon §45 Compactness in Metric Spaces in Topology by J.R. Munkres.
