Bounded and Closed but not Compact I had been proposed to construct not compact set that is however bounded and closed.
I could easily imagine from the different metric - such as discrete metric where 
$d(x,y) =0$ if $x=y$ and $d(x,y) =1$ if $x \neq y$ then M itself is closed since it is metric itself and bounded by $D(x,2)$, but is not compact since there's $D(x,1/2)$ which is open cover of $M$ with no finite sub-cover 
any how to construct bounded and closed but not compact set in complete metric?

A subset of a banach space is compact if and only if it is closed and totally bounded. The closed unit ball in $l^1(\mathbb R)$ centered around $0$ is not totally bounded. For example because it is not a finite union of open balls of radius $\frac{1}{2}$, simply because each such ball can contain at most one of the sequences of the form $0,0,\dots, 0, 1,\dots $ (this is clear by the triangle inequality).
