It is proven that if two operators $\hat{X}$ and $\hat{Y}$ commute, then the multiplication of them will be hermitian, i.e. if $\hat{X}\hat{Y}=\hat{Y}\hat{X}$, then $\left(\hat{X}\hat{Y}\right)^\dagger=\hat{X}\hat{Y}$.
My question is that is the opposite true as well? In other words, if the multiplication of two operators is hermitian, then will they commute? If yes, I need a proof.
The statement of your assertion is a little off, but essentially yes. The correct statement is that two Hermitian operators must commute if their product is also Hermitian. The proof is entirely straightforward as a Hermitian product implies $XY=(XY)^\dagger$ but $$ (XY)^\dagger=Y^\dagger X^\dagger=YX $$ using that $X$ and $Y$ are both Hermitian themselves. Hence, $XY=YX$.
So, in fact the full statement of the theorem would be given two Hermitian operators $X$ and $Y$, the operators commute if and only if their product is also Hermitian.