There is a question in my mind which seems to be silly but I am desperately wanting the answer.
Why a metric is defined from $X\times X$ to $\mathbb R$ and not to the set of nonnegative reals? I mean if we consider the function from $X\times X$ to $\mathbb R^+\cup \{0\}$ and drop the axiom $d(x,y)\ge 0$, then will it work?
Please through some light in this context.
Yes, it definitely will work.
Moreover, metric is sometimes defined as a map $d \colon X \times X \to \mathbb{R}$ satisfying:
$d(x, y) = 0 \Leftrightarrow x = y,\ \forall x, y \in X.$
$d(x, y) = d(y, x),\ \forall x, y \in X.$
$d(x, z) \leqslant d(x, y) + d(y, z),\ \forall x, y, z \in X.$
We can deduce $d(x, y) \geqslant 0$ from these axioms as follows. Take $z = x$ in the third axiom obtaining:
$$\forall x, y \in X:\ 0 = d(x, x) \leqslant d(x, y) + d(y, x) = 2d(x, y) \Rightarrow d(x, y) \geqslant 0$$
So this axiom can be omited but is ussualy included into the axiom list probably for the reason stated in Starlight's comment. Another reason for this can be introducing the generalization of the notion of a metric (e.g. pseudometric, quasimetric and others) without adding new axioms but just omiting some of the given ones (in that case non-negativity can't be deduced from other axioms so it must be included into the list of axioms).