I am not calling this gravitational time dilation because that is a relativistic effect due to the equivalence principle. ... But GR predicts it only through accelerating frames and thus bringing in relativity. I do not see any relativity in my case. Its just spacetime curvature.
What you have is gravitational time dilation. But even if you don't call it such, in any static, curved spacetime with time dilation between stationary clocks, at least some of the clocks are necessarily accelerated. A static spacetime has a privileged timelike direction in which geometry remains the same, and its metric tensor $g$ can therefore be put in the following form:
$$\mathrm{d}s^2 = -e^{2\phi}\,\mathrm{d}t^2 + h_{ij}\,\mathrm{d}x^i\,\mathrm{d}x^j\text{,}$$
where $\phi$ is a function that may depend on any spatial coordinates $x^i$ ($i = 1,2,3$), but not on $t$. A clock stationary in this spacetime will have four-acceleration components $a^\alpha = g^{\alpha\beta}\phi_{,\beta}$. If there is some time dilation between stationary clocks at different positions, then $\phi$ is not constant with respect to the spatial coordinates, and so at least some of the partial derivatives $\phi_{,\beta}$ don't vanish: therefore, some (perhaps all) of the stationary clocks are necessarily accelerated.
Furthermore, any theory in which test particles in freefall take geodesics in a curved spacetime automatically satisfies the weak equivalence principle.
GR explains it in the following way:- Suppose you are falling from space into the earth, in a higher position your speed is not very high and thus you experience less time dilation (as time dilation directly depends on your velocity) while when you get closer to the surface, your speed becomes higher and you experience more time dilation.
No. Consider two observers stationary in a gravitational field, one on top of another, separated by a height along which gravity doesn't change much (so we can approximate the gravitational field as uniform). Since they are accelerated, the situation is equivalent to having them outside a gravitational field but in a vertically accelerating rocket.
Let's say the rocket is transparent. Then a horizontal light beam entering the accelerating rocket will be see to curve downward in the rocket's frame. Thus it must do so in the case of a gravitational field as well. With a few light beam reflections, this situation is identical your setup.
Won't this also lead to the slowing of time in a higher gravitational area (where the force of gravity is more)?
Yes.
Can this be an actual reason for gravitational time dilation or not?
Sort of. Instead of passing the light beam horizontally, it would have been much simpler to pass it vertically. Suppose two observers stationary in a gravitational field, one on top of each other, separated by a height small enough so that gravity doesn't change much over it, pass each other light pulses at regular intervals (according to their reckoning). This situation is then equivalent to doing the same outside a gravitational field but in a vertically accelerating rocket. Then it becomes obvious that the bottom is receiving signals at a rate different from which the top is sending them, as the bottom is catching up to the light pulses while they are in transit.
Again: this isn't really specific to GTR. The argument would be virtually identical in any curved, static spacetime, because stationary observers in such a spacetime are accelerated. We can then consider the situation in a local inertial frame and everything else follows as above. The key relativistic insight is simply that gravitational fields disappear in inertial frames.
Even GR predicts that a clock will run slower in high gravity but doesn't actually give a reason.
The usual presentation of GTR's reasoning regarding gravitational time dilation involves a setup morally equivalent to yours, except most often presented in a simpler way, because having light rays go in the same direction as the acceleration is easier. It has nothing to do with speed gained while falling as you seem to think.
You have written that the bottom is receiving signals at a different rate. But I would like to remind you that gravity does not cause any speed of change in light when light is vertical. Please explain.
The only difference is the orientation of the light clock. The speed of light is the same in every inertial frame, and in the inertial frame instantaneously comoving with the light clock (free-falling from rest), both ends of the light clock accelerate upward. Thus, the bottom accelerates toward a downward pulse and the top accelerates away from an upward pulse, so the distance those pulses travel is different.
As for how the situation looks in the stationary frame, the coordinate speed of light is different at the top of the light clock is different from the coordinate speed of light at the bottom.With an explicit radial coordinate, the the above metric (without $c=1$ units) is
$$\mathrm{d}s^2 = -e^{2\phi}\,c^2\,\mathrm{d}t^2 + h_{rr}\,\mathrm{d}r^2 + \{\text{other spatial terms}\}\text{,}$$
so that vertical light has
$$\left|\frac{\mathrm{d}r}{\mathrm{d}t}\right| = c\frac{e^{\phi}}{\sqrt{h_{rr}}}\text{.}$$
Since top part has proper time $\tau_\text{top}$ with $\mathrm{d}\tau_\text{top} = e^{\phi_\text{top}}\,\mathrm{d}t$ and the bottom part has proper time $\tau_\text{bot}$ with $\mathrm{d}\tau_\text{bot} = e^{\phi_\text{bot}}\,\mathrm{d}t$, usually with $\phi_\text{top}\neq\phi_\text{bot}$, they will disagree with the amount of time vertical light pulses take to traverse the same $r$-coordinate interval.
By the way, GTR predicts that for weak fields (appropriate near our Earth), $h_{rr} \approx e^{-\phi}$ and $\phi\approx\Phi/c^2$, where $\Phi$ is the Newtonian gravitational potential. But the above keeps the discussion general to any static spacetime, whether it agrees with what GTR predicts for the situation or not.
I did not get how is the two observers case similar to one in a rocket outside space as you have said.
In a local inertial frame, the two observers accelerate upward. The condition that the gravitational field does not change much over the relevant height means that we can consider their acceleration to be the same. That's the acceleration of the rocket, but you can forget about the rocket if you wish and just consider the same acceleration in a local inertial frame.
