If $C$ is a proper class, then there is a surjection $s : C \to \textbf{On}$. Here, On is the proper class of ordinals.
My idea for such a surjection is to show that the class $\{ \text{rank}(a) : a \in C \}$ is exactly the same as the class of ordinals. I'm pretty sure the idea is right, but I'm not sure about how to show this.
The idea isn't exactly right. Consider, for example, the class $C$ of limit ordinals - $C$ has no element of rank (say) $17$.
However, you're moving in the right direction. If $C$ is a proper class, then the class of ranks of elements of $C$ forms an unbounded class of ordinals; can you see how to surject an unbounded class of ordinals onto all the ordinals? (HINT: there's a very nice recursive construction here. First think about how you'd surject an arbitrary infinite set of finite ordinals onto all of $\omega$ ...)