You can prove this without looking at any of the specific cases by doing a first-order perturbation of the differential equation that defines the time-evolution operator.
We start with
$$
i \hbar \frac{\partial}{\partial t} U(t, t_0) = H U(t, t_0).
$$
Or, rearranging some terms,
$$
\frac{\partial}{\partial t} U(t, t_0) = - \frac{i}{\hbar} H U(t, t_0).
$$
At a time $t + \delta t$, the above equation tells us that to first order in $\delta t$ we have
$$
U(t + \delta t, t_0) = U(t, t_0) - \frac{i}{\hbar} H U(t, t_0) \delta t.
$$
If $U$ is unitary, then we have
$$
I = U^{\dagger} U (t + \delta t, t_0) = \left[ U^{\dagger}(t, t_0) + \frac{i}{\hbar} U^{\dagger}(t, t_0) H^{\dagger} \delta t\right] \left[ U(t, t_0) - \frac{i}{\hbar} H U(t, t_0) \delta t \right].
$$
Expanding this and keeping only terms to first order in $\delta t$ yields
$$
I = U^{\dagger} U (t, t_0) - \frac{i}{\hbar} U^{\dagger} (t, t_0) H U (t, t_0) \delta t + \frac{i}{\hbar} U^{\dagger}(t, t_0) H^{\dagger} U (t, t_0) \delta t.
$$
Since we demanded that $U$ is unitary for all time, we should also have $U^{\dagger} U (t, t_0) = I.$ Substituting this and canceling $I$ from both sides of the equation yields
$$
0 = \frac{i}{\hbar} U^{\dagger}(t, t_0) \left(H^{\dagger} - H \right) U(t, t_0) \delta t.
$$
We must therefore have $H^{\dagger} = H$. So we have shown that if $U$ is unitary, then $H$ must be Hermitian.
For the other direction, we return to our first order expansion:
$$
U^{\dagger} U (t + \delta t, t_0) = U^{\dagger} U (t, t_0) - \frac{i}{\hbar} U^{\dagger} (t, t_0) H U (t, t_0) \delta t + \frac{i}{\hbar} U^{\dagger}(t, t_0) H^{\dagger} U (t, t_0) \delta t.
$$
If $H$ is Hermitian, then
$$
U^{\dagger} U (t + \delta t, t_0) = U^{\dagger} U(t, t_0).
$$
So $U^{\dagger} U(t, t_0)$ is constant for all $t$. Since $U(t_0, t_0)$ is the identity, we must have $U^{\dagger} U(t, t_0) = I$ for all $t$. We have thus proven that if $H$ is Hermitian, then $U$ must be unitary.
Best Answer
One proof of eq. (3): Everything commute$^1$ by definition under the time (normal, radial, etc) ordering symbol: $$T\left([A,B]\right)~=~0,\tag{A}$$ so BCH formula simplifies $$ T\left(e^Ae^B\right)~=~T\left(e^{A+B+\frac{1}{2}[A,B]+[\text{nested commutator terms}]}\right)~=~ T\left(e^{A+B}\right) ,\tag{B}$$ which proves OP's sought-for eq. (3).
Another proof of eq. (3): Use the Trotter formula.
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$^1$ The main point is that the time ordering procedure $T$ does not take operators to operators, but symbols/functions to operators, cf. this & this related Phys.SE post and links therein.