Proof that the degree of every vertice in a Minimum Spanning Tree (MST) is constant (i.e. not dependent on the number of vertices $n$) Given a set of points S in $\mathbb{R}^2$ we have a MST of these points. I want to proof that every vertice has a degree in $\mathcal{O}(1)$, i.e. it does not depend on the number of vertices in the MST. Intuitively i can see that this has to be true since otherwise there would be cycles and a lot of edges and therefore probably not a MST. Any hint is appreciated!

Hint: Suppose we have a minimum spanning tree $T$, take a vertex $v$, with two other vertices $v_1,v_2$ adjacent to $v$, and suppose the angle $\angle v_1 v v_2$ is $\alpha$.

Since $v$ is adjacent to $v_1$ and $v_2$ in the MST, we must have $d(v,v_1) + d(v,v_2)$ less than both of the alternatives $d(v,v_1) + d(v_1,v_2)$ or $d(v,v_2) + d(v_1,v_2)$. Therefore, we just need to argue that (if $\alpha$ is small enough) we'll have that $d(v_1,v_2)$ is smaller than the maximum of $d(v,v_1)$ and $d(v,v_2)$.