Prove that if all costs are proportional to distances, then a shortest tour cannot intersect itself.
If replacing two of the intersecting edge with two others edge that pairing up with the same node, it will give a shorter tour. Would that be enough for the proof?
To answer your exact question: No, it is not enough to note that the replacement edges are shorter than the removed edges (which I assume you've proven using the triangle relation). But you're on the right track!
You also need to prove that the resulting graph is still a tour; the obvious objection being that it is easy to think of examples where the result of this edge swapping is the union of two disjoint tours.
Your language is a bit unclear, but I think we would agree that there are two choices for "two other edges that pairing up in the same node". Can you make an argument that it is always the case that that one of those choices still yields a tour?