A reflective subcategory $C$ of $D$ is a full subcategory of $D$ such that for any object $d\in D$, if $d$ is isomorphic to some $c\in C$, then $d\in D$ and furthermore the inclusion $i:C\rightarrow D$ has a left adjoint $r:D \rightarrow C$.
I am trying to show:
Let $C$ be a reflective subcategory of $D$. I am trying to show that if $D$ is co-complete then so is $C$.
I have come up with a solution, but it does not rely on the above introduced definition, so I am wondering if it is correct.
Let $J:I\rightarrow C$ be a diagram. Then $i\circ J: I \rightarrow D$ is a diagram and since $D$ is co-complete we have that $colim_{I} J$ exists in $D$. But then since we have that $r$ is left adjoint to $i$, it preserves colimits so, $$r(\text{colim} _I (i\circ J)) \cong \text{colim}_I (r\circ i\circ J)\cong \text{colim}_I (J)$$
since we have that $r\circ i \Rightarrow 1_C$. Hence we have exhibited the colimit as an object of $C$.
Is this a correct proof?
The definition of reflective subcategory only asks for the inclusion functor to be a right adjoint. So it makes sense that you're only using the assumption that $i$ is a right adjoint.
Your argument seems okay. Maybe for clarity one could say something about the last step: you need to use that $ri$ is naturally isomorphic to the identity functor. This follows from the fact that $i$ is fully faithful (see Lemma 4.5.13 of Riehl's book Category Theory in Context).