Let's stay within the 2D case for now. A standard proof is constructive:
(1) Pick an arced edge on the boundary of the circle. Note its labeling (for example, (1, 2)).
(2) Walk into the circle through your chosen edge into a simplex of the triangle.
(3) If the simplex carries three different labels, then two of them most be antipodal, so Tucker's Lemma is satisfied.
(4) If instead the simplex is entirely labelled with 1's and 2's, then walk through the (1, 2) edge that you didn't enter from, and repeat step (3) or (4) on your new simplex.
(5) Eventually, you will either dead-end in a tri-labeled simplex in the middle of the triangle, or you will walk out of the circle through another arced edge. If you leave the circle, then you've eliminated two edges; pick a new edge and try again.
(6) Since there are an even number of vertices on the boundary of the circle, there are an odd number of edges on the boundary of the circle. So eventually, one of your paths must dead-end inside the circle.
This proof does not rely on the fact that the triangulation of the circle is antipodal symmetric; instead, it only relies on the fact that there are an even number of vertices on the boundary of the circle. So why do we require antipodal symmetry, when the weaker "even number of edges" condition implies the same conclusion?
Statement $(6)$ is false: the number of edges on the boundary is equal to the number of vertices on the boundary, so both are even.