I do find a solution to a related question but it does not address Delaunay triangulation explicitly.
If you define the Delaunay triangulation for a set $P$ of points in the plane as a triangulation such that no point in $P$ is inside the circumcircle of any triangle in the triangulation, then a simple proof is given by lifting the points to the paraboloid $z=x^2+y^2$. A simple calculation shows that a point is inside the circle defined by three points iff the lifted point is below the plane defined by the three lifted points. Hence, the projection of the lower convex hull of the lifted points is the Delaunay triangulation of the original points.
So, if you believe that the convex hull is (or can be) triangulated, then you believe that a Delaunay triangulation exists.
Figure from the book Geometry and Topology for Mesh Generation by Edelsbrunner.