In class we saw that if a functor $F:\mathbb C \to \mathbb D$ is an equivalence (with quasi-inverse $G$), it must be full and faithful. Can I prove it the following way?
Fixed $C,C'$ objects of $\mathbb C$, one has: $$\mathrm{Hom}_\mathbb C(C,C')\xrightarrow F \mathrm{Hom}_\mathbb D(FC,FC')\xrightarrow G \mathrm{Hom}_\mathbb C(GFC,GFC');$$ since $GF$ induces a bijection between $\mathrm{Hom}_\mathbb C(C,C') $ and $ \mathrm{Hom}_\mathbb C(GFC,GFC')$, $G$ must be full and $F$ must be faithful. By analogy, using $FG\cong 1_\mathbb D$, one finds that $F$ is full and $G$ faithful too.
The proof we saw in class follows a similar argument, but uses more category theory (mine relies basically set theory): am I losing information somewhere or this proof, even if it's not very constructive, is correct?
The proof is OK, but a small detail is missing in your argument. Namely, you have to prove that if $F \cong \mathrm{id}_{\mathcal{C}}$ for a functor $F : \mathcal{C} \to \mathcal{C}$, then $F$ is fully faithful (apply this to $GF$ in your notation). More generally, if $F \cong G$ for two functors $F,G : \mathcal{C} \to \mathcal{D}$, then $F$ is fully faithful if (and only if) $G$ is fully faithful. The proof is, of course, straight forward. If $\alpha : F \to G$ is an isomorphism and $f,g : X \rightrightarrows Y$ are morphisms in $\mathcal{C}$, we see that
$\begin{align*} F(f)=F(g) & \implies \alpha_Y \circ F(f) = \alpha_Y \circ F(g) \\ & \implies G(f) \circ \alpha_X = G(g) \circ \alpha_X \\ & \implies G(f) = G(g) \\ & \implies f = g. \end{align*}$