We need to show that \[\begin{equation} \int \psi_1^*(L^2 \psi_2)d\Omega=\left[\int \psi_2^*(L^2 \psi_1)d\Omega\right]^*, \tag{1} \end{equation}\] where \[\begin{equation} L^2=-\left(\frac{1}{\sin \theta} \frac{\partial}{\partial \theta}\sin\theta\frac{\partial}{\partial \theta}+ \frac{1}{\sin^2 \theta}\frac{\partial^2}{\partial \phi^2}\right). \tag{2} \end{equation}\] The second part of \(L^2\) operator is easier to handle. The relevant part of the integral is the \(\phi\) integral, which can be computed as follows \[\begin{eqnarray} \int_0^{2 \pi} \psi_1^*\left[\frac{1}{\sin^2 \theta}\frac{\partial^2}{\partial \phi^2}\psi_2\right]d\phi&=&\int_0^{2 \pi} \frac{\partial}{\partial \phi}\left[\psi_1^*\frac{1}{\sin^2 \theta}\frac{\partial}{\partial \phi}\psi_2\right]d\phi-\int_0^{2 \pi} \left(\frac{\partial}{\partial \phi}\psi_1\right)^*\left(\frac{1}{\sin^2 \theta}\frac{\partial}{\partial \phi}\psi_2\right)d\phi\nonumber\\ &=&-\int_0^{2 \pi} \frac{\partial}{\partial \phi}\left[\left(\frac{\partial}{\partial \phi}\psi_1\right)^*\frac{1}{\sin^2 \theta}\psi_2\right]d\phi+\int_0^{2 \pi} \left(\frac{1}{\sin^2 \theta}\frac{\partial^2}{\partial \phi^2}\psi_1\right)^*\psi_2d\phi\\ &=&\left[\int_0^{2 \pi} \left(\frac{1}{\sin^2 \theta}\frac{\partial^2}{\partial \phi^2}\psi_1\right)\psi_2^*d\phi\right]^*, \tag{3} \end{eqnarray}\] where we dropped the first terms in the first two lines as the they are the difference of the integrand at \(\phi=2\pi\) and \(\phi=0\), and that is zero as \(\phi\) coordinate is \(2\pi\) periodic.
The first part of \(L^2\) operator seems to be harder because when we integrate by parts \(\frac{\partial}{\partial \theta}\) will act on \(\sin\theta\), which will complicate the problem. However, we can avoid it by a change of variable \(u=\cos\theta\). The relevant part of the integral is the \(d cos\theta\) integral, and with the above transformation it becomes,
\[\begin{eqnarray} \int_0^{ \pi} \psi_1^*\frac{1}{\sin \theta} \frac{\partial}{\partial \theta}\left(\sin\theta\frac{\partial}{\partial \theta}\psi_2\right)d\cos\theta&=&\int_{-1}^{1}\psi_1^*\frac{\partial}{\partial u}\left((1-u^2)\frac{\partial}{\partial u}\psi_2\right)du\nonumber\\ &=&-\int_{-1}^{1}\frac{\partial}{\partial u}\psi_1^*\left((1-u^2)\frac{\partial}{\partial u}\psi_2\right)du\nonumber\\ &=&-\int_{-1}^{1}\left((1-u^2)\frac{\partial}{\partial u}\psi_1^*\right)\frac{\partial}{\partial u}\psi_2du\nonumber\\ &=&\int_{-1}^{1}\frac{\partial}{\partial u}\left((1-u^2)\frac{\partial}{\partial u}\psi_1^*\right)\psi_2du\nonumber\\ &=&\left[\int_{-1}^{1}\psi_2^*\frac{\partial}{\partial u}\left((1-u^2)\frac{\partial}{\partial u}\psi_1\right)du\right]^*\nonumber\\ &=&\left[\int_0^{ \pi} \psi_2^*\frac{1}{\sin \theta} \frac{\partial}{\partial \theta}\left(\sin\theta\frac{\partial}{\partial \theta}\psi_1\right)d\cos\theta\right]^*, \tag{4} \end{eqnarray}\] where we dropped again some surface terms as \(u^2-1=0\) at \(u=\pm1\). (If you prefer \(\sin \theta d\theta\) integral instead of \(d cos\theta\) integral, you will not need to change the variable.) This completes the proof that \(L^2\) is Hermitian.