Proving that a mapping is bijective 
The proof I'm trying to write is for the problem above. I'm kind of sure that it's right because it feels right but it also feels kind of easy, so it sort of feels wrong? Anyways, I just need confirmation that I'm not being an idiot. 

Proof Attempt:
Let $\phi: F \to F$ be a mapping defined such that $\phi(\alpha) = \overline{\alpha}$. To prove that this is a homomorphism of fields, we need to show that it preserves sums and products. So, let $\alpha_1,\alpha_2 \in F$. Then:
$\phi(\alpha_1 \cdot \alpha_2) = \overline{\alpha_1 \cdot \alpha_2}$
$\phi(\alpha_1 \cdot \alpha_2) = (r_1r_2-s_1s_2)-i(r_1s_2-r_2s_1)$
$\phi(\alpha_1 \cdot \alpha_2) = (r_1-is_1)(r_2-is_2) = \overline{\alpha_1} \cdot \overline{\alpha_2}$
$\phi(\alpha_1 \cdot \alpha_2) = \phi(\alpha_1) \cdot \phi(\alpha_2)$
Then, we also have:
$\phi(\alpha_1+\alpha_2) = \overline{\alpha_1+\alpha_2} = (r_1+r_2)-i(s_1+s_2) = (r_1-is_1)+(r_2-is_2) = \overline{\alpha_1} +\overline{\alpha_2}$
$\phi(\alpha_1+\alpha_2) = \phi(\alpha_1) + \phi(\alpha_2)$.  
This proves that it is a homomorphism of fields. 
Now, we need to prove that this mapping is bijective. 
First, we prove injectivity. Let $\alpha_1,\alpha_2 \in F$ such that $\alpha_1 = r_1+is_1$ and $\alpha_2 = r_2+is_2$, where $r_1,r_2,s_1,s_2 \in \mathbb{Q}$. Then:
$\phi(\alpha_1) = \phi(\alpha_2)$
$\implies \overline{\alpha_1} = \overline{\alpha_2}$
$\implies r_1-is_1 = r_2-is_2$
$\implies (r_1 = r_2) \land (s_1 = s_2)$
$\iff \alpha_1 = \alpha_2$
Secondly, we prove surjectivity. Clearly, $\phi(F) \subset F$. Let $\alpha_1 \in F$. Then, $\overline{\alpha_1} \in F$. So:
$\phi(\overline{\alpha_1}) = \alpha_1$
$\implies \alpha_1 \in \phi(F)$. 
This shows that $F \subset \phi(F)$. Hence, $\phi(F) = F$. Since $\phi$ is surjective and injective, it follows that it is bijective and an automorphism. 
Does the argument above make sense?

Your proof is fine. For the proof of bijectivity, you can simply show that $\phi^{-1} = \phi$, so since $\phi$ has a two-sided inverse, it is bijective.
