We first prove orthogonality of non-degenerate eigenvectors of the Hamiltonian. Consider the braket and act with the Hamiltonian in both directions,
$ \left\langle\alpha | H |\beta\right\rangle = E_\alpha\left\langle\alpha |\beta\right\rangle = E _\beta\left\langle\alpha |\beta\right\rangle $
If the states are not orthogonal ($\left\langle\alpha |\beta\right\rangle \neq 0 $) then we would get a contradiction since we assume the states are non-degenerate ($E_\alpha\neq E_\beta $). So we must have
$\left\langle\alpha |\beta\right\rangle = 0 $
for distinct states.
2. Now we need to prove that the braket of two eigenstates is equal to $1$ up to a phase. Consider the braket:
$ \left\langle\alpha |\alpha\right\rangle = \sum_n \left\langle\alpha |n\right\rangle \left\langle n |\alpha\right\rangle = \left\langle\alpha |\alpha\right\rangle \left\langle\alpha |\alpha\right\rangle $
where we have inserted a sum over the states of the Hamiltonian and then used the orthogonality relation that we proved above. Now we can divide both sides by $\left\langle\alpha |\alpha\right\rangle $ to get
$\left\langle\alpha |\alpha\right\rangle = 1 $