If $p$ is a prime and $p \neq 2,3,$ show that $\bar{p} = \bar{1}$ or $\bar{p} = \bar{5}$ in $\mathbb{Z}_6.$
First, I showed that $\bar{p} \neq \bar{0},\bar{2},\bar{3},\bar{4}.$
(1) $\bar{p} \neq \bar{0},$ because $6|p - 0 \implies 6|p \implies$ $p$ is not prime.
(2) $\bar{p} \neq \bar{2},$ because $6|p - 2$ implies that $6$ divides some odd number $x$ (all primes $p > 2$ are necessarily odd, and so $p - 2$ is odd), which in turn implies that $2$ divides $x,$ which contradicts that $x$ is odd.
(3) $\bar{p} \neq \bar{3},$ because $6|p - 3$ implies that $6d = p - 3$ for some integer $d,$ and $6d + 3 = 3(2d + 1) = p,$ which contradicts that $p$ is prime.
(4) $\bar{p} \neq \bar{4}$ for the same reason $\bar{p} \neq \bar{2}.$
Now, I simply show that $\bar{p} = \bar{1}$ and $\bar{p} = \bar{5}$ are each possible. I give the examples:
(1) $p = 7 \implies \bar{p} = \bar{1}$
(2) $p = 11 \implies \bar{p} = \bar{5}.$
Your proof is correct, but you can dispense with providing examples for $\bar1$ and $\bar5$. Since $\bar p$ can only be one of $\bar0$ to $\bar5$ – a complete residue system – by eliminating four of those cases you have shown that it must be one of the remaining two.