Hint: First prove that $iLc$ satisfies the condition in 3, where $i:D\to C$ is the inclusion. Then use apply 2 to conclude.
Regarding "localization" the point is that $L$ is characterized by the arrows it inverts, so the language is being imported from commutative algebra. Specifically, the motivating situation is that of $R[S^{-1}]$-modules, where $S$ is a multiplicative set in a commutative ring $R$. Every $R[S^{-1}]$ is functorially an $R$-module by restricting the scalars, and this functor is fully faithful, with the left adjoint $R[S^{-1}]\otimes_R (-)$. So this is a reflective subcategory. Furthermore an $R$-module admits an $R[S^{-1}]$-action if and only if it is local for those $R$-module maps inverted by tensoring with $R[S^{-1}]$, as in Riehl's point 3. The classical algebraic framework would reduce those maps to the maps $s:R\to R$ determined by elements of $S$, while the general categorical framework instead asks for locality with respect to all maps $\eta_A:A\to A[S^{-1}]$, for $A$ an $R$-module. We can bridge these frameworks by observing that locality with respect to each $s$ is equivalent to locality with respect to the single unit map $R\to R[S^{-1}]$, which implies locality with respect to every $\eta_A$ by considering the action of $L$ on a presentation of $A$.
Let me try to understand what you said with your $X,Y,A,A',B$, and then I'll give an example to see what's happening.
You take $B\in Y$, and a reflection of it in $X$, say $h: B\to A$. If $f: A\to A'$ is not in $X$, you get $f\circ h : B\to A'$. You say "this must factor through some $A\to A'$ which is in $X$; and since you chose $f$ not to be in $X$, this seems impossible.
But the point is that there will be some $f' : A\to A'$ such that $f'\circ h = f\circ h$, with $f'\in X$. And this $f'$ will be unique in $X$.
An easy example of a non full reflective subcategory is, given any category $C$ with products, the diagonal $\Delta: C\to C^2$.
We may clearly see it as a subcategory: the subcategory on objets $(A,A)$ and morphisms $(f,f)$ between those. In general, it's not full : there will, in general, be objects $A,B$ with two distinct morphisms $f,g:A\to B$ (if $C$ is not a poset, you're guaranteed to find such $A,B$), and so $(f,g) : (A,A)\to (B,B)$ is not in the image of $C$.
A left adjoint to this inclusion is given by the coproduct $(A,B)\mapsto A\coprod B$ (and if you want to really view it as a subcategory, it's $(A,B)\mapsto (A\coprod B, A\coprod B)$)
It's clear that this is not a full reflective subcategory, as if it were, then the reflector applied to $(A,A)$ would be just $(A,A)$ : here it's $(A\coprod A, A\coprod A)$ which is in general wildly different.
Now in my example, what's happening with your question : take $f,g: A\to B$ that are different, so we get $(f,g) : (A,A)\to (B,B)$ which is not in $\Delta(C)$, and suppose $(A,A)$ is the reflection of some $(E,F)$, that is $A= E\coprod F$.
Then $f,g$ are determined by $f_0,g_0: E\to B$ and $f_1,g_1: F\to B$, and the map $(E,F)\to (B,B)$ is given by $(f_0,g_1)$. But then its reflection is $([f_0,g_1],[f_0,g_1]) : (A,A)\to (B,B)$, which is different from what we started with, i.e. $(f,g) = ([f_0,f_1],[g_0,g_1])$ .
So we see that we do get a different f$'$ which will be in $X$, and which will be the only map in $X$ to satisfy $f'\circ h = f\circ h$.
(where for maps $h : E\to D, k:F\to D$, I let $[h,k]: E\coprod F\to D$ denote the uniquely determiend map)
Best Answer
Sure, for instance take $\mathbf{C}\subset \mathbf{Set}$ the full subcategory consisting of the cardinal sets. In general, if $\mathbf{C}\subset \mathbf{D}$ is a reflexive subcategory, then any skeleton of $\mathbf{C}$ is also reflexive in $\mathbf{D}$.