Why does SAT-UNSAT $\in NP \implies NP = coNP$
Suppose that $X$ is a coNP problem which is NP-hard.
Let $A$ be any problem in NP. Since $X$ is NP-hard, there is a polytime reduction from $A$ to $X$. Since $X$ is in coNP, this shows that $A$ is in coNP.
We have shown that $\mathsf{NP} \subseteq \mathsf{coNP}$. Now suppose $B$ is in coNP. Then $\overline{B}$ is in NP, and so in coNP. Hence $B$ is in NP. This shows that also $\mathsf{coNP} \subseteq \mathsf{NP}$, and so $\mathsf{NP} = \mathsf{coNP}$.