Do minimum spanning trees drawn on points in $\mathbb{R}^2$ always have non-intersecting edges?

Question

Preamble

This question is motivated by the question "Embedded minimum spanning trees for visualizing effects of dimensionality reduction?".

Suppose I were to begin with a collection of points $X = \{x_1, \cdots, x_n \} \subset \mathbb{R}^2$, and computed a minimum spanning tree on these points with edges drawn as the straight line segments.

I know that trees are always planar graphs, but that doesn't mean that all choices of embedding don't have intersections of the line segments representing the edges.

Example

Below is a sample from a pair of IID standard normal data variables. I've empirically inducted that the resulting MST always has a planar embedding from looking at many such plots.

Example Code

from itertools import combinations

import matplotlib.pyplot as plt
import networkx as nx
import numpy as np

X = np.random.normal(size=1000).reshape((500,2))
pos = {i:X[i] for i in range(X.shape[0])}

g = nx.Graph()
for comb in combinations(range(X.shape[0]),r=2):
    g.add_edge(*comb, weight=np.linalg.norm(X[comb[0]] - X[comb[1]]))

g = nx.minimum_spanning_tree(g)

nx.draw(g, pos, node_size=20)
plt.show()

Question

Are the line segments drawn on the plane guaranteed to be non-intersecting?

Answer 1

This is a summary of the discussion in the comments posted as an answer:

The Eucledean MST is always a subgraph of the Delaunay triangulation. This can be proved easily by arguing with contradiction:

If an edge $e$ between points $A$ and $B$ of the EMST does not belong in the Delaunay triangulation then, by considering the circle $C(e)$ with diameter $e$, we could find another point $D$ inside $C(E)$. This allows us to replace the edge $e$ with another suitably chosen edge $e'$ in a way that the total length of the edges decreases, a contradiction.

Now since the Delaunay triangulation is an embedded planar graph the same holds for the EMST.