Your proof seems to work, but I'm not sure. However you can use lambda calculus to show a beta equivalence. That works guaranteed: $$I:=\lambda x.x, \quad K:=\lambda x. \lambda y. x, \quad S := \lambda x. \lambda y. \lambda z. xz(yz)$$ The first term reduces to: $$ SII = \lambda z. (\lambda x.x) z ((\lambda x.x)z) \rightarrow_\beta \lambda z.zz \\ S(K(SII)) = \lambda y . \lambda z. ( \lambda y . \lambda z . zz) z (yz) \rightarrow_\beta \lambda y . \lambda z. yz(yz)$$ And the second to (Keep in mind to rename conflicting variables!): $$ SS = \lambda y . \lambda z. ( \lambda x. \lambda y . \lambda z. xz(yz))z(yz) \rightarrow_\beta \lambda y . \lambda z. (\lambda z'. z z' (yz z'))\\ SSI = \lambda z . \lambda z' . z z' (z z') =_\alpha \lambda y . \lambda z. yz(yz)$$ Therefore $SII$ and $S(K(SII))$ are $\beta$-equivalent.

Hope this helps :)

Your proof seems to work, but I'm not sure. However you can use lambda calculus to show a beta equivalence. That works guaranteed: $$I:=\lambda x.x, \quad K:=\lambda x. \lambda y. x, \quad S := \lambda x. \lambda y. \lambda z. xz(yz)$$ The first term reduces to: $$ SII = \lambda z. (\lambda x.x) z ((\lambda x.x)z) \rightarrow_\beta \lambda z.zz \\ S(K(SII)) = \lambda y . \lambda z. ( \lambda y . \lambda z . zz) z (yz) \rightarrow_\beta \lambda y . \lambda z. yz(yz)$$ And the second to (Keep in mind to rename conflicting variables!): $$ SS = \lambda y . \lambda z. ( \lambda x. \lambda y . \lambda z. xz(yz))z(yz) \rightarrow_\beta \lambda y . \lambda z. (\lambda z'. z z' (yz z'))\\ SSI = \lambda z . \lambda z' . z z' (z z') =_\alpha \lambda y . \lambda z. yz(yz)$$ Therefore $SSI$ and $S(K(SII))$ are $\beta$-equivalent.

Hope this helps :)