How can we prove that the Hamiltonian for any quantum system is Hermitian? By applying partial time derivative to
$$\psi_t \rightarrow U \psi_{t_0}$$
we end up with an expression for the Hamiltonian
$$H = i\hbar\frac{\partial U_{t}}{\partial t}U^{\dagger}_{t}$$
where $U$ is the unitary time evolution operator. 
How can we show that $(\partial U_{t} / \partial t) U^\dagger_t$ is anti-Hermitian? If $(\partial U_t / \partial t)U^\dagger_t$ is anti-Hermtian surely we can multiply $i$ to  $(\partial U_t / \partial t)U^\dagger_t$ which would make $i(\partial U_t/\partial t)U^\dagger_t$ Hermitian. Right?
 A: Since $U_t$ is unitary, we have:
$$U_t U_t^\dagger = \mathbb I$$
Taking a derivative from both sides:
$$\frac{\partial U_t}{\partial t} U_t^\dagger +U_t (\frac{\partial U_t}{\partial t})^\dagger =0$$
so:
$$\frac{\partial U_t}{\partial t} U_t^\dagger =-U_t (\frac{\partial U_t}{\partial t})^\dagger$$
or:
$$\frac{\partial U_t}{\partial t} U_t^\dagger = -\Big(\frac{\partial U_t}{\partial t} U_t^\dagger \Big)^\dagger$$
Meaning that $\frac{\partial U_t}{\partial t} U_t^\dagger$ is anti-Hermitian.
EDIT:
The OP brings up an interesting question in the comments: Can any anti-Hermitian operator-valued function $A(t)$ be written in the form $\dot{U}(t) U^\dagger(t)$ for some unitary operator $U(t)$?
The answer is yes, as long as $A(t)$ is differentiable. To see this, let's define the operator $U(t)$ through the following initial value problem:
$$\dot{U}(t) = A(t) U(t) \qquad \text{with} \quad U(0) = \mathbb I \qquad (*)$$
First of all, taking the Hermitian conjugate of $(*)$ we get:
$$\dot{U}^\dagger(t) = -U^\dagger(t)A(t)  \qquad \text{with} \quad U^\dagger(0) = \mathbb I \qquad (**)$$
Using $(*),(**)$ we can show that $U(t)$ is indeed unitary. Let's calculate the derivative of $U^\dagger(t) U(t)$:
$$\frac{\partial }{\partial t} \Big(U^\dagger(t) U(t) \Big) =  \dot{U}^\dagger(t)  U(t) +U^\dagger(t) \dot{U}(t) = -U^\dagger (t) A(t) U(t)+U^\dagger(t) A(t) U(t) = 0$$
So $U^\dagger(t) U(t)$ is constant in time meaning that:
$$U^\dagger(t) U(t)= U^\dagger(0) U(0) = \mathbb I$$
Therefore, $U(t)$ is indeed unitary. Now from $(*)$ we can easily write:
$$A(t) = \dot{U}(t) U^\dagger(t)$$
meaning that we've proved our claim. 
Second EDIT:
Emilio Pisanty points out in the comments that our defining initial value problem for $U$ may not have a solution in the general case of an infinite dimensional Hilbert space. The above argument obviously breaks down for those settings.
