Given a category $D$ with subcategory $C$ it is known that the inclusion functor $i: C \longrightarrow D$ reflects isomorphisms if $C$ is a full subcategory. I'm not sure about the necessity of the full subcategory condition here.
Given a morphism $f$ in $C$ we obtain the morphism $i(f) =f$ in $D$. I'm not sure why it doesn't now follow that if $f$ is an isomorphism in $D$, and since $f$ is also in $C$, then $f$ must be an isomorphism in $C$? Maybe it doesn't follow because the inverse of $f$ in $D$ may not be in $C$?
Yes, this is exactly the problem, and you can use this observation to construct the universal counterexample, as follows. Let $D$ be the "free category on two isomorphic objects"; that is, it consists of two objects $c, d$ and two non-identity morphisms $f : c \to d, f^{-1} : d \to c$ such that $f \circ f^{-1} = \text{id}_d, f^{-1} \circ f = \text{id}_c$. Now take $C$ to be the subcategory containing $c, d$, and $f$, but not $f^{-1}$.
The real lesson is that non-full "subcategories" are not really subcategories at all, because non-fullness allows the morphisms to differ so drastically they might as well be entirely different objects. An inclusion of a non-full subcategory is really just a faithful functor $F : C \to D$ and this produces examples like the following: we can take $D = \text{Set}$, $C$ to be the category $\text{Top}$ of topological spaces, and $F : \text{Top} \to \text{Set}$ to be the forgetful functor. In $\text{Top}$ there are continuous bijections which are not homeomorphisms, such as the map
$$[0, 2\pi) \ni t \mapsto e^{it} \in S^1$$
or examples involving discrete or indiscrete topologies; these correspond to morphisms $f : c \to d$ in $\text{Top}$ such that $F(f)$ is an isomorphism but $f$ is not. In this example it's a little silly to say that $\text{Top}$ is a "subcategory" of $\text{Set}$ even though it's possible to rearrange this example a bit so that $F$ is injective on objects (by replacing $\text{Set}$ with the equivalent category whose objects are topological spaces but whose morphisms are arbitrary functions, ignoring continuity). $\text{Top}$ is just a quite different category entirely.