Can quantum information conservation verified in this specific situation? Consider a projective measurement of a superposition of states:
$$\frac{1}{\sqrt{2}} |\Psi_1\rangle + \frac{1}{\sqrt{2}} |\Psi_2\rangle \xrightarrow[measurement]{projective} |\Psi_1\rangle, \qquad (|\Psi_1\rangle \neq |\Psi_2\rangle)$$
Assuming an unitary evolution of the "universe = quantum system + measure apparatus" the state evolves as:
$$\left(\frac{1}{\sqrt{2}} |\Psi_1\rangle + \frac{1}{\sqrt{2}} |\Psi_2\rangle\right)\otimes |\chi_0\rangle \xrightarrow[measurement]{unitary} |\Psi_1\rangle \otimes |\chi_1\rangle$$
But:

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*Can we somehow prove that the initial state information is contained in the final state, or is it just an assumption we make, difficult to prove in practice?

*Would it be possible to construct a very simple measuring device where this would be easier to verify?

 A: 
Can we somehow prove that the initial state information is contained in the final state, or is it just an assumption we make, difficult to prove in practice?

Not only can we not prove this, but it is not true. If the final state is indeed $|\Psi_1 \rangle$, as it would be in textbook Quantum Mechanics, then there is no way of inferring the initial state from the final state. After all, any superposition with a non-zero coefficient in front of $|\Psi_1 \rangle$ could have led to the final state $|\Psi_1 \rangle$. Thus there are infinitely many initial states consistent with just 1 final state.

Would it be possible to construct a very simple measuring device where this would be easier to verify?

There is no way to verify that unitary evolution would give you the non-superposition state $|\Psi_1 \rangle$. That is because unitary evolution does not give you that state; it is inconsistent with the linearity of the Schrödinger equation. Here is how we know:
First, we know that measurement of an eigenstate just gives you back that eigenstate. If we assume that measurement happens through unitary evolution, then
$$|\Psi_1 \rangle \otimes |\chi_0 \rangle \xrightarrow[measurement]{unitary} |\Psi_1 \rangle \otimes |\chi_1 \rangle $$
and
$$|\Psi_2 \rangle \otimes |\chi_0 \rangle \xrightarrow[measurement]{unitary} |\Psi_2 \rangle \otimes |\chi_2 \rangle $$
Then, since the unitary evolution in quantum mechanics requires the sum of solutions to be a solution, we can add those solutions together to find out that
$$\alpha |\Psi_1 \rangle \otimes |\chi_0 \rangle + \beta |\Psi_2 \rangle \otimes |\chi_0 \rangle\xrightarrow[measurement]{unitary}\alpha |\Psi_1 \rangle \otimes |\chi_1 \rangle + \beta |\Psi_2 \rangle \otimes |\chi_2\rangle $$
Taking the example of $\alpha = \beta = \frac{1}{\sqrt{2}}$, this reads
$$ \left(\frac{1}{\sqrt{2}}|\Psi_1 \rangle  +  \frac{1}{\sqrt{2}}|\Psi_2 \rangle \right)\otimes |\chi_0 \rangle\xrightarrow[measurement]{unitary}\frac{1}{\sqrt{2}} |\Psi_1 \rangle \otimes |\chi_1 \rangle + \frac{1}{\sqrt{2}}|\Psi_2 \rangle \otimes |\chi_2 \rangle $$
...which is in direct contradiction with the statement you wanted to prove, which I'll copy below:
$$\left(\frac{1}{\sqrt{2}} |\Psi_1\rangle + \frac{1}{\sqrt{2}} |\Psi_2\rangle\right)\otimes |\chi_0\rangle \xrightarrow[measurement]{unitary} |\Psi_1\rangle \otimes |\chi_1\rangle$$
