$\overline M = M$ (closure of $M$ is $M$) I have an intuition that an entire metric space (not a subset of it) does not have boundaries, so its closure would be itself, but how to prove it?
 A: I gather from your comments that you define the closure $\overline{A}$ of a set $A\subseteq M$ to be the set of all points in $M$ which are adherent points of $A$. I assume your definition of adherent point agrees with the one here.
Note that for any set $A\subseteq M$, $\overline{A}\subseteq M$, by definition.
Also, for any set $A\subseteq M$, $A\subseteq \overline{A}$. Proof: Let $x\in A$. For every open set $U$ containing $x$, $U$ contains a point of $A$ (namely $x$). So $x$ is an adherent point of $A$, and $x\in \overline{A}$.
Taking $A = M$, $\overline{M}\subseteq M$ and $M\subseteq \overline{M}$, so $M = \overline{M}$.
A: We know that the distance of a point $x$ to a set $A$ is $d(x,A)=\inf_{y\in A}d(x,y)$. Now using your definition, $$\overline{M}=\{x\in M\,|\,d(x,M)=0\}$$ So clearly, $\overline M\subseteq M$.
If $x\in M$ then $d(x,M)=0$, thus $x\in \overline M$. 
So $M=\overline M$
A: Let $M$ be any metric space, let $S$ be any subset of $M$ (possibly $M$ itself), let $x \in M$.  
Definition 1: We say that $x$ is an adherent point of $S$ if for any open set $V$ of $M$ with $x \in S$, the intersection $S \cap V$ is nonempty.
Definition 2:  The closure of $S$ in $M$, denoted $\overline{S}$, is the set of adherent points of $S$.
We always have $\overline{S} \subseteq M$.  This is clear, because points of $\overline{S}$ are defined to be certain points of $M$.
Lemma: For any subset $S$ of $M$, $S \subseteq \overline{S}$.
Proof: Let $x \in S$.  We want to show that $x$ is an adherent point of $S$.  Let $V$ be any open subset of $M$ with $x \in V$.  We need to argue that the intersection $S \cap V$ is nonempty.  But this is clear, because $x$ is in this intersection.  $\blacksquare$
Now let $S = M$.  We want to show that $\overline{M}$ is equal to $M$.  By the remark after Definition 2, we know that $\overline{M} \subseteq M$.  But on the other hand, the Lemma tells us that $M \subseteq \overline{M}$.  So $M = \overline{M}$.
A: Observe that given any topology $T$ on $M$ , $\emptyset$ and the whole space $M$ always lie in $T$. This  just follows from the definition of topology.
Now since sets in $T$ are defined to be $open$ $sets$ and complement of open sets are $closed$, the proof follows. 
A: $\overline{M}$ by your definition is a closed set of $M$, so $\overline{M} \subseteq M$.  At the same time, any subset $E$ of $M$ is contained in its closure $\overline{E}$.  So $M \subseteq \overline{M}$.  Hence $M = \overline{M}$.
A: For $M\subset\overline{M}$:
$\forall x\in M$, $\forall B$ an open ball containing $x$, we have $x\in B\cap M$, then $B\cap M\ne\emptyset$. By the definition of "closure", we obtain that $x\in\overline{M}$. Thus $M\subset\overline{M}$.
For $\overline{M}\subset M$:
$\forall x\in\overline{M}$, by the definition of "closure", we konw that $\exists B_x$ the open ball containing $x$, such that $B_x\cap M\ne\emptyset$. Since $B_x$ is an open ball with respect to the entire metric space $M$, it is thereby a relatively open set in $M$, which implies $B_x\subset M$. By recalling $x\in B_x$, we have $x\in M$. Hence $\overline{M}\subset M$.
We are done.
