Let $\ m\ge3$, and let $\ a_i$ be the natural numbers less than or equal to $\ m$ that are coprime to $\ m$ put in the following order: $$\ a_1<a_2<\cdots<a_\frac{\phi(m)}{2}\le \frac{m}{2}\le a_{\frac{\phi(m)}{2}+1}<a_{\frac{\phi(m)}{2}+2}<\cdots<a_{\phi(m)}.$$
If $\ a_{\frac{\phi(m)}{2}}>\frac{m}{2}$ and $\ a_{\frac{\phi(m)}{2}+1}\ge\frac{m}{2}$ then $\ a_{\frac{\phi(m)}{2}}+a_{\frac{\phi(m)}{2}+1}>m$ which is wrong.
If $\ a_{\frac{\phi(m)}{2}}\le\frac{m}{2}$ and $\ a_{\frac{\phi(m)}{2}+1}<\frac{m}{2}$ then $\ a_{\frac{\phi(m)}{2}}+a_{\frac{\phi(m)}{2}+1}<m$ which is wrong.
If $\ a_{\frac{\phi(m)}{2}}>\frac{m}{2}$ and $\ a_{\frac{\phi(m)}{2}+1}<\frac{m}{2}$ then $\ a_{\frac{\phi(m)}{2}+1}<a_{\frac{\phi(m)}{2}}$ which is wrong.
So $\ a_{\frac{\phi(m)}{2}}>\frac{m}{2}$ or $\ a_{\frac{\phi(m)}{2}+1}<\frac{m}{2}$ is wrong, $\ a_{\frac{\phi(m)}{2}}\le\frac{m}{2}$ and $\ a_{\frac{\phi(m)}{2}+1}\ge\frac{m}{2}$ is true, and it gives the result.
Does this proof work?
Your proof is correct, but you should clearly indicate where the proof starts and that you are using the result on the sum of two symmetric elements in the proof.