The counit of a reflection is an isomorphism

Since $\mathcal{A}$ is a full subcategory of $\mathcal{C}$, [L4] gives $\mathcal{A}(A,B)=\mathcal{C}(A,B)=\mathcal{C}(I(A),I(B))$ for all $A,B\in \mathcal{A}$, and the inclusion $I$ acts on those hom-sets as the identity; so every $I_{A,B}$ is bijective and [L1] makes $I$ fully faithful. By its surjectivity there is ${\delta }_A:A\to RI(A)$ with $I({\delta }_A)={\eta }_{I(A)}$, and by its injectivity that ${\delta }_A$ is unique. The first triangle identity in [L2] gives $I({\epsilon }_A\circ {\delta }_A)=I({\epsilon }_A)\circ {\eta }_{I(A)}=1_{I(A)}=I(1_A)$, so faithfulness gives ${\epsilon }_A\circ {\delta }_A=1_A$.

Naturality of the counit at ${\delta }_A$ gives ${\delta }_A\circ {\epsilon }_A={\epsilon }_{RI(A)}\circ RI({\delta }_A)$. Since $I({\delta }_A)={\eta }_{I(A)}$, functoriality gives $RI({\delta }_A)=R({\eta }_{I(A)})$, and the second triangle identity in [L2] makes the right side $1_{RI(A)}$. Thus ${\delta }_A$ is a two-sided inverse of ${\epsilon }_A$, so [L3] proves the claim.