prove that every complete graph with 4 or more vertices has two spanning trees with disjoint edges

I read a possible proof of "every complete graph with 4 or more vertices has two spanning trees with disjoint edges" in the answer of another question.
That is, first claim that every complete graph with 4 or more vertices has a wheel as its subgraph, which I can understand.
Then claim that every wheel will have 2 spanning trees with disjointed edges because one is complement graph of another. I can find 2 spanning trees with disjointed edges in some specific wheels, but how to prove it generally?
Or is there other ways to prove the statement "every complete graph with 4 or more vertices has two spanning trees with disjoint edges"? Thanks.

A wheel with $$n$$ vertices is a cycle $$C_{n-1}$$, joined to a single vertex $$v$$. Label the vertices of this cycle $$u_1, u_2, \dots, u_{n-1}$$.
Let the first spanning tree $$T_1$$ be the path $$(v, u_1, u_{n-1}, u_{n-2}, \dots, u_2)$$. Let the second spanning tree $$T_2$$ be the 'star' centered at $$v$$, with edges $$vu_2, vu_3, \dots, vu_{n-1}$$, as well as the single edge $$u_1u_2$$ of the cycle.
In the diagram below, $$T_1$$ is blue, and $$T_2$$ is red.