proof by contradiction:
assume S is a shortest path tree of G (with the same root), and that there exists a vertex v where dist(v) from Dijkstra's algorithm does not equal its distance in S.
Lets examine the two options:
the dist(v) from Dijkstra's algorithm on G is bigger than its value in S. If so this implies that Dijkstra's algorithm is wrong, since there is a shorter path to this vertex.
the dist(v) from Dijkstra's algorithm on G is smaller than its value in S - this means that S could not be a valid shortest path tree, because a shorter path exists to vertex v, thus contradicting our initial assumption.
Q.E.D.