Proving that the hermitian conjugate of the product of two operators is the product of the two hermitian congugate operators in opposite order

As leftaroundabout wrote, integration by parts is unuseful. You don't have the expressions for operators, so there is no reasoning for it. But you may use following: \begin{align} \langle \Psi_{1}|(\hat {A}\hat {B})^{+} |\Psi_{2}\rangle & = \langle \Psi_{2} | \hat {A}\hat {B} |\Psi_{1}\rangle^{*} \\ & = \sum_{c}\langle \Psi_{2}| \hat {A}| c \rangle^{*} \langle c|\hat {B}| \Psi_{1}\rangle^{*} \\ & = \sum_{c}\langle c| \hat {A}^{+}| \Psi_{2} \rangle \langle \Psi_{1}|\hat {B}^{+}| c\rangle \\ & =\sum_{c}\langle \Psi_{1}| \hat {B}^{+}| c \rangle \langle c|\hat {A}^{+}| \Psi_{2}\rangle \\ & = \langle \Psi_{1}|\hat {B}^{+}\hat {A}^{+} |\Psi_{2}\rangle , \end{align} where I used definition of hermitian conjugate, $$ \langle \Psi_{1}| \hat {A}^{+}|\Psi_{2}\rangle = \langle \Psi_{2}| \hat {A}|\Psi_{1}\rangle^{*}, $$ and basis $|c\rangle $ of eigenvectors of an operator in a Hilbert space, $\langle c| c\rangle = 1$; $\sum_c|c\rangle\langle c|=\mathbb 1$

You don't actually need to pick a basis as indicated in Andrew McAdams's answer.

This is easiest to prove in mathy notation (as opposed to Dirac notation) where $(\cdot, \cdot)$ is the inner product, then for all vectors $\phi$ and $\psi$ in the Hilbert space, and for operators $A$ and $B$, we have \begin{align} (\phi, AB\psi) = (A^\dagger\phi, B\psi) = (B^\dagger A^\dagger\phi, \psi) \end{align} while on the other hand \begin{align} (\phi, AB\psi) = ((AB)^\dagger\phi, \psi) \end{align} which implies $B^\dagger A^\dagger = (AB)^\dagger$ as desired.