As the question states, we are to show that $C = \{ f : f \text{ is a function} \}$ is a proper class.
Assume for the sake of contradiction that $C$ is a set. Then consider the class function $\Phi$ which takes as input a function $f$ and spits out the set of inputs to $f$. So if $f : A \to B$ then $\Phi(f) = A$. By replacement schema, $\Phi[C]$ is a set. However, since $\{ S : S \text { is a set} \}$ is not a set, we have a contradiction.
Does this proof make sense?
I understand what you're trying, but it's not obvious that $\Phi (C) = \{S: S \text{ is a set}\}$.
Try this. If $C$ is a set, then by Comprehension, $I=\{f \in C~|~ \exists x (f= \{\langle x, x \rangle \}) \}$ is a set. If $I$ is a set, then by Replacement, $S=\{ \text{dom}(f) ~|~ f \in I \}$ is a set. But $S$ is the set of all sets, and we know that doesn't exist.