Clearly, such a mapping is injective, hence by invariance of domain the image is open. From isometry, it also easily follows the image is closed. Since $\mathbb{R}^2$ is connected and the image is non-empty, the map is surjective. :P
I am working on some slides for a presentation to some highschool students about isometries. And I realized that the book I learned about isometries from assumed surjectivity.
I was just at a loss because I have so far presented isometries as something they already knew and were familiar with vis rotations, translations, and reflections from elementary school.
But the surjectivity part seemed to be something you couldn't think of an isometry without, but I couldn't get something that fit nicely with my presentation. I didn't want to go to great lengths just to prove surjectivity.