I may have a proof but I am not sure.
Let $A_1 \subseteq A_2 \subseteq B$ be a chain of coreflective subcategories. Let $b$ be an object of $B$. Then $b$ has a coreflection $a_2 \xrightarrow{c_2} b$ with $a_2 \in Obj(A_2)$, and $a_2$ itself has a coreflection $a_1 \xrightarrow{c_1} a_2$ with $a_1 \in Obj(A_1)$. Consider an arbitrary morphism $f : a'_1 \rightarrow b$ with $a'_1 \in Obj(A_1)$. By assumption $a'_1$ is also an object of $A_2$ so there exists a unique $f' : a'_1 \rightarrow a_2$ such that $f = c_2 \circ f'$. By coreflectiveness of $A_1$ in $A_2$, the morphism $f'$ can also be factored $f' = c_1 \circ f''$ for a unique $f''$. Then $(c_1 \circ c_2) \circ f'' = f$. To prove uniqueness of $f''$ for the latter equality, let $g : a'_1 \rightarrow a_1$ be another such morphism. Then $c_2 \circ (c_1 \circ g) = c_2 \circ f'= f$ so $c_1 \circ g = f'$ by uniqueness of $f'$. Then $c_1 \circ g = c_1 \circ f''$ so $g = f''$ by uniqueness of $f''$.
Your proof is correct. Note that this is just a specific case of a more general fact : the composition of two left adjoint is left adjoint to the composition of their right adjoints. In other words : given functors $F_1:\mathcal{X}\to \mathcal{Y}$, $F_2:\mathcal{Y}\to \mathcal{Z}$, $G_1:\mathcal{Y}\to \mathcal{X}$ and $G_2:\mathcal{Z}\to \mathcal{Y}$ such that $F_1\dashv G_1$ and $F_2\dashv G_2$, then $F_2F_1\dashv G_1G_2$. The proof is obtained by simply composing the natural bijections : $$\operatorname{Hom}_{\mathcal{Z}}(F_2F_1(X),Z)\cong \operatorname{Hom}_{\mathcal{Y}}(F_1(X),G_2(Z))\cong \operatorname{Hom}_{\mathcal{X}}(X,G_1G_2(Z)).$$
In your case $F_1$ and $F_2$ are the two inclusion functors; and in fact your proof itself is basically the particularisation of the composition I wrote above to your specific case.