My conclusion is that the Symmetry never breaks
Your conclusion is wrong. Let's say (1) labels the event when both observers start off at rest, and (2) labels the event when, at the end of the experiment, they come back together at rest. The proper times recorded on the clocks then are ( earth's gravity disregarded for simplicity ) :

and

The metric tensor for the traveller is isomorphic, but not identical, to the Minkowski metric, and will be of the general form

This is straightforward during the outbound and inbound segments of the journey, but potentially very non-trivial at the turn-around point. The exact form of the acceleration function there will explicitly depend on the geometry of the world line at that region, i.e. on how exactly the turn-around is executed. However, at no point during the turn-around phase does the acceleration function vanish.

It is easy to see that these metric tensors are

**not** the same for the two observers, hence the two frames are

**not** symmetric, contrary to your conclusion. The only times when these frames are symmetric is when they are both in relative

**inertial** motion ( in which case the metric tensor becomes the Minkowski tensor in both frames ).

While the Earth will observe rocket’s time elapsed less, if you had been on the rocket, you will observe the Earth’s time elapsed less.
This is incorrect. For this to be the case, the two proper time integrals above would need to be the same - which they obviously aren't.

The Symmetry has never broken
Yes it was. See above.