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Suppose in an empty space two rocketships exists with person A and B of same age inside each. Let A move with velocity c/2 away from B and return. They'll find out that B grew older while A didn't have that change in age due to relative passing of time. But this incident can also be interpreted in another way such that, we can assume B to be moving away while A stood still as there is no other frame to compare with. How will we define proper time and improper time in this case? Do we have to take the initial point of A in space into consideration? Doesn't this mean we're taking space as a universal reference frame?

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This problem was always a struggle for me, and the solution is that acceleration is measured with a scale; that is, acceleration is measured by measuring a force. Thus, the twin who in the rocket ship feels an acceleration as they speed up to their maximum velocity, and more importantly, they feel (and measure) an acceleration when turning around to come back. It is always the twin that feels the acceleration that ages more slowly, and it is clear from this definition of acceleration who that twin is.

Kraigolas
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You just realized that constant speed is symmetrically relative. This means that there is a paradox called the twin paradox. Both twins could see the other speed away and age less.

Now the solution is in SR the change of reference frames, but the real solution is GR.

As per GR, and the equivalence principle, when the spaceship of A is returning, it needs to turn around.

Now turning around, needs deceleration and acceleration. Acceleration is the same as being in a gravitational field (equivalence principle). Acceleration is absolute.

Now when the spaceship of A turns around, it accelerates and during that acceleration, A is slowing down in the time dimension (relative to B), so he ages less during the turn around.

There is a age difference after the turn around, and that age difference is kept while A returns (with constant speed) to B.

Thus, when they meet again, A will seem to have aged less.