I need to find an example of a functor $F:\mathcal A\to\mathcal B$ such that $F$ is faithful but there exist distinct maps $f_1$ and $f_2$ with $F(f_1)=F(f_2)$. Is the following example correct?
Let $(A=\{a_1,a_2,a_1',a_2'\},\leq)$ be preordered set, where $\leq$ is defined with $a_1\leq a_2, a_1'\leq a_2'$ and let $(B=\{b_1,b_2\},\leq)$ be a preordered set, where $\leq$ is defined with $b_1\leq b_2$. We regard $(A,\leq)$ as category $\mathcal A$ and $(B,\leq)$ as category $\mathcal B$ in usual way. We take $f_1$ to be $a_1\leq a_2$ and $f_2$ to be $a_1'\leq a_2'$, and we define $F$ with $F(a_1)=F(a_1')=b_1$, $F(a_2)=F(a_2')=b_2$ and $F(f_1)=F(f_2)=b_1\leq b_2$.
Yes, that is correct. A faithful functor is injective when restricted to morphisms from a given pair (domain, codomain), but it's possible that two morphisms with different domain/codomain are sent to the same morphism. It's because a faithful functor isn't required to be injective on objects.
If you want an even simpler example, consider the category $\mathcal{C}$ with two objects, $x$ and $y$, and two morphisms: the identity of $x$ and the identity of $y$. Then consider the category $\mathcal{D}$ with only one object $z$ and one morphism, the identity of $z$. Finally consider the only functor $F : \mathcal{C} \to \mathcal{D}$. Then $F(\operatorname{id}_x) = F(\operatorname{id}_y)$, even though $F$ is faithful.