A commutative or cocommutative Hopf algebra has involutive antipode

Assume $H$ is commutative. Applying $S$ to the left antipode equation $\sum S(h_{(1)})h_{(2)}=\epsilon (h)1$ and using anti-multiplicativity gives $\sum S(h_{(2)})S^2(h_{(1)})=S(\epsilon (h)1)=\epsilon (h)S(1)=\epsilon (h)1$, where $S(1)=1$ is [F2]. Commutativity of $H$ lets us reorder each product, so $S^2\ast S=u\epsilon$. Since also $S\ast \mathrm{i}\mathrm{d}=u\epsilon$, associativity gives $S^2=S^2\ast (S\ast \mathrm{i}\mathrm{d})=(S^2\ast S)\ast \mathrm{i}\mathrm{d}=\mathrm{i}\mathrm{d}$.

Assume instead that $H$ is cocommutative. Applying $S$ to the left antipode equation as in step 1.1 gives $\sum S(h_{(2)})S^2(h_{(1)})=S(\epsilon (h)1)=\epsilon (h)S(1)=\epsilon (h)1$ by [F2]; cocommutativity permits interchange of the two Sweedler indices, yielding $S\ast S^2=u\epsilon$. Since $\mathrm{i}\mathrm{d}\ast S=u\epsilon$, associativity gives $S^2=(\mathrm{i}\mathrm{d}\ast S)\ast S^2=\mathrm{i}\mathrm{d}\ast (S\ast S^2)=\mathrm{i}\mathrm{d}$.

The commutative calculation proves the first sufficient branch, and the cocommutative calculation proves the second; either hypothesis therefore implies $S^2={\mathrm{i}\mathrm{d}}_H$. No finite-dimensionality or choice principle is used.