@LastIronStar: Do you have Roman's book at your disposal?

@LastIronStar Very good.

Let me try to help you so that you yourself can answer your question. More specifically the following,

For that let's go back at the definition of the Subcategories given in Roman's book.

For the definition to work, you will first need to show that $F\mathcal{C}$ is a category. Have you done that @LastIronStar?

@LastIronStar No it's not. Note that in Roman's definition he writes, "Let $\mathcal{C}$ be a category. A subcategory $\mathcal{D}$ of $\mathcal{C}$ is a category for..." So unless you show that $F\mathcal{C}$ is a category itself, how can you apply Roman's definition of a subcategory to conclude that $F\mathcal{C}$ is a subcategory of $D$?

@LastIronStar Wait, by this comment did you mean the following question,

or the following,

Sorry, I misunderstood.

If you want to show that $F\mathcal{C}$ is a category then you need to ensure that for all $A,B,C\in\text{obj}(F\mathcal{C})$ and for $i\in\text{hom}_{F\mathcal{C}}(A,B),j\in\text{hom}_{F\mathcal{C}}(B,C)$ there exists a morphism $j\circ i$ such that $j\circ i\in\text{hom}_{F\mathcal{C}}(A,C)$. Isn't it? Now take a look at your own question, do you get the answer @LastIronStar?

@LastIronStar Exactly. In other words if $F(j\circ i)\ne Fj\circ Fi$, right?

I will leave the rest to you. I think you can handle it now.

Just forgot to mention one thing. Before reading Roman's book, I think it is better if you listen to his lectures and then complement it with his book (I myself did the same) @LastIronStar.

@LastIronStar See this.