Monomorphisms are preserved by Representable Functors Emily Riehl's "Category Theory in Context", ${\rm Exercise}~2.1.{\rm ii}.$

Prove that if $F:{\rm C}\to{\rm Set}$ is representable, then $F$ preserves monomorphisms, i.e., sends every monomorphim in ${\rm C}$ to an injective function. Use the contrapositive to find a covariant set-valued functor defind on your favorite concrete category that is not representable. 

I have a rough idea how to approach, but $a)$ I am not sure if my reasoning is valid and $b)$ how to justify one step. Anyway, lets start. 
Keep in mind that $f$ is a monomorphism if it is left
cancellable, i.e $f\circ x=f\circ y\implies x=y$. Given a representation of $F:{\rm C}\to{\rm Set}$ by an object $c$ and thus a natural isomorphism $F\cong{\rm Hom}(c,-)$ it follows that
\begin{align*}
f\circ x&=f\circ y\tag1\\
F(f\circ x)&=F(f\circ y)\tag2\\
\cong~{\rm Hom}(c,f\circ x)&={\rm Hom}(c,f\circ y)\tag3\\
{\rm Hom}(c,x)&={\rm Hom}(c,y)\tag{$\color{red}4$}\\
\cong F(x)&=F(y)\tag5
\end{align*}
Line $(3)$ and line $(5)$ follow from the given natural isomorphism. What bothers me is line $(4)$ (as emphasized above). I am not entirely sure if one can argumentate like this. As $f$ is a monomorphism this property is inherited to hom-sets involving $f$. Anyway, I do not know how to proper phrase this step (if it's even possible to do so!).

Is my proof correct, if so how to justify line $(4)$; if not, please explain what went wrong. Additionally, I would like to see an example for the second part of the exercise as I cannot really think of one in particular.

Thanks in advance!
 A: I think you went too deep following a wrong track of reasoning, and now are very much confused as to what objects you are even dealing with. There isn't much to the problem if you approach it correctly.
Suppose $f: X \to Y$ is mono. We want to show that $F(f): F(X) \to F(Y)$ is injective function of sets. Since $F$ is representable, $F$ is naturally isomorphic to $\operatorname{Hom}(A, -)$ for some $A \in \operatorname{Ob}(\mathbf{C})$. This means that we have the following commutative diagram in the category of sets:
$\require{AMScd}$
\begin{CD}
F(X) @>{F(f)}>> F(Y)\\
@V{\mu_X}VV @V{\mu_Y}VV\\
\operatorname{Hom}(A, X) @>{f \circ -}>> \operatorname{Hom}(A, Y)
\end{CD}
where $\mu_X, \mu_Y$ are isomorphisms (bijections), and we need to prove that $F(f)$ is injective. Since the vertical arrows are bijections, to prove that the top arrow is injective, it's enough to prove that the bottom arrow is injective. What it means for bottom arrow to be injective? Take any $x, y \in \operatorname{Hom}(A, X)$. For the bottom arrow $f \circ -$ to be injective, we need to have that whenever $f \circ x = f \circ y$, necessarily $x = y$. But that follows immediately from $f$ being monomorphism.
