How can you conclude that gravity is a conservative force? A force field $F_i(x)$ is conservative if for every curve $C$ from a point $y_1$ to a point $y_2$, we have $\int\limits_C F_i(x)\mathrm{d}x^i$, so that the energy difference between $y_1$ and $y_2$ is independent of the curve taken from one to the other. Equivalently, the integral around a closed curve must be zero, $\oint\limits_C F_i(x)\mathrm{d}x^i=0$ for every closed curve $C$.
This is the definition of conservative force. Okay I agree but What I cannot understand is How did you verify that between two point in a gravitational field the work done in moving a object from one point to another is independent of the path, I mean there are literally infinite numbers of path that we can have between those two point so How can we say that it is independent of path ?
How did you reach to the conclusion that gravity is a conservative force ?
 A: The definition of a conservative force came after the observation of conservation of energy and the accumulation of data for the gravitational field.

How did you reach to the conclusion that gravity is a conservative force ?

First one accepts conservation of energy, and the proof that in a gravitational field the potential energy of an object is fixed by its position. Look at hydroelectricity. No matter  which path the water has taken to enter  the dam, the energy it can give is fixed by the dam height to the hydroelectric plant.
One sees this by mathematically modeling all the different paths an object can take to be found at a height h with the fixed potential energy.

A conservative force depends only on the position of the object. If a force is conservative, it is possible to assign a numerical value for the potential at any point and conversely, when an object moves from one location to another, the force changes the potential energy of the object by an amount that does not depend on the path taken, contributing to the mechanical energy and the overall conservation of energy. If the force is not conservative, then defining a scalar potential is not possible, because taking different paths would lead to conflicting potential differences between the start and end points.

A: Just show that $F_idx^i$ is total derivative, i.e. $F_idx^i=dW$. Then the path integral reduces to $\int_C dW$, which is independent of the path and depends only on the initial and final point.
So:
$$F_idx^i=-G\frac{mM}{r^3}\left(xdx+ydy+zdz\right)=-G\frac{mM}{2r^3}d\left(x^2+y^2+z^2\right)=-G\frac{mM}{r^2}dr.$$ As this is of the form $f(r)dr$, it is indeed a total derivative.
A: Are you looking for a mathematical proof (which has been given by others), or an experimental demonstration?
If gravity is not conservative then that means there would two paths up a mountain that take different amounts of energy to ascend (friction excluded).
So if you started on a bike at the top, you could free-wheel down the high-energy path, then coast up the low-energy path and when you got back to the top, you'd still have some excess energy (you'd still be moving). You could go round again and get even faster. And again and again, gaining free energy all the time.
Can you see what other conservation law you're breaking here?
A: Stokes' theorem tells us that for any vector field, the closed line integral of that field is equal to the surface integral of the curl of that field over any surface bounded by the closed loop. In this case, for a gravitational field
$$\oint \vec{g}\cdot d\vec{l} = \int (\nabla \times \vec{g})\cdot d\vec{A}.$$
Clearly, the LHS of this equation would be the work done (per unit mass) in moving an object around a closed loop in a gravitational field.
But
$$\vec{g} = \frac{GM}{r^2} \hat{r}$$
in spherical coordinates, where $\hat{r}$ is a unit vector in the radial direction. Taking the curl of this field
in spherical coordinates, then because there are no $\theta$ or $\phi$ components, and $g_r \neq f(\theta, \phi$),
$$\nabla \times \vec{g}=0.$$
Given that, then the RHS of Stokes' theorem is always zero and so the work done by the gravitational field around a closed path is always zero.
Note that the same argument applies to any central, symmetric force
A: The force field due to a small element of mass (which we can think of as a point mass) is spherically symmetric and central, which makes it a conservative field. For the case of field due to a point mass, consider resolving each tiny segment of your path into 2 components, one along the radial direction and one along the circular direction. Work is only done when you travel along the radial components of the path, because that's where the force is along the path. No work is done along the circular component of the path because force is perpendicular to the displacement.
Now you can take any path to get from point A to point B, but no matter what path you take the radial components of the displacement and their corresponding force along that displacement will be the same (because of the central and spherically symmetric nature of the field), and we can hence conclude that the change in potential energy does not depend on the path.
The gravitational field of a continuous object is just the vector sum of all of the fields due to the individual mass elements, and is therefore also conservative even though the two fields (of a point mass and of an extended body) may look quite different. 
