If $NP \neq co-NP$ then no $NP-complete$ problem can be in $co-NP$
We will show the contrapositive: if some NP-complete problem is in coNP, then NP=coNP. Suppose that $L$ is some NP-complete problem which is in coNP. Let $A$ be any problem in NP. Then there is a polytime reduction $f$ from $A$ to $L$ (since $L$ is NP-complete), which means that $x \in A$ iff $f(x) \in L$. Since $L$ is in coNP, this shows that $A$ is in coNP. Hence NP is a subset of coNP.

Now let $B$ be any problem in coNP. Then there is a polytime reduction $f$ such that $x \in B$ iff $f(x) \in \overline{L}$ (since $L$ is NP-complete and $\overline{B}$ is in NP). Since $\overline{L}$ is in NP, this shows that $B$ is in NP, and so coNP is a subset of NP. We conclude that NP=coNP.