Proving that if coNP $\neq$ NP then P $\neq$ NP
It is maybe easier to consider the contrapositive, that is ${\sf P}={\sf NP} \Rightarrow {\sf NP}={\sf coNP}$.

So assume ${\sf P}={\sf NP}$, then 


for every $L\in {\sf NP}$, we have $L\in {\sf P}$, and since the languages in ${\sf P}$ are closed under complement,  $\bar L\in {\sf P}$ and therefore $L\in {\sf coNP}$.
for every $L\in {\sf coNP}$, we have $\bar L\in {\sf P}$, and since the languages in ${\sf P}$ are closed under complement,  $ L\in {\sf P}$ and therefore $ L\in {\sf NP}$.


Remark: Note that if ${\sf P}={\sf NP}$ the polynomial time hierarchy collapses to the lowest level, which implies that ${\sf P}={\sf NP}={\sf coNP}={\sf PH}$.