Why $G/C_G(a) leq G/zeta(G)$ in this lemma?

I am sorry, but the answer of Chris Custer is not correct. You have $zeta(G) subseteq C_G(a) subseteq G$, and this only implies that $G/C_G(a) cong (G/zeta(G))/(C_G(a)/zeta(G))$. Probably you need divisibility and locally finiteness along the line to show that $G/C_G(a)$ is isomorphic to a direct summand of $G/zeta(G)$.

Note added later Aha! Now I see the full proof. So indeed the statement you are questioning is totally wrong. As I pointed out above, $G/C_G(a)$ is a quotient of a divisible group hence divisible. That is what is needed. And of course that a bounded divisible group is trivial. So thanks for adding the full proof and good question you asked! I already +1-ed your post.