Only one unique adiabatic path between two states
For an infinitesimal adiabatic process $dS_\mathrm{surroundings} = 0$ since no heat is exchanged with the surroundings. But $dS_\mathrm{system} = 0$  is only true for a reversible process, since only then  $dS_\mathrm{system} = dq/T$. For an irreversible process $dS_\mathrm{system} \neq dq/T =0$. Therefore
$$dS_{\mathrm{irrev}}\neq dS_{\mathrm{rev}}=0$$
This means that the state accessed with a reversible process through an integral change $\int dS_{\mathrm{rev}}=0$ cannot be accessed directly through an irreversible adiabatic process, for which $\Delta S = \int dS_{\mathrm{irrev}} \neq 0$.
Now since S is a state function, the terminal states of the two processes must therefore differ (QED).
As regards the second part (how do we know which is the reversible), you can use the approach just outlined to conclude that for the reversible process $\Delta S =0$.