To prove that (iii) implies (i), assume w.l.o.g. that $\tau\neq\sigma$. We first need to show that $\tau\subset \partial \sigma$. If this is not the case then there exists a hyperplane $H$ containing $\tau$ such that $H$ is not a supporting hyperplane for $\sigma$. Then there are $x,y\in\sigma$ on different sides of $H$, which may be chosen in $N$ as $\sigma$ is rational, so that $x+y\in\tau$, contadicting (iii).
Thus $\tau\subset \partial \sigma$. Let $\sigma'$ be the minimal face of $\sigma$ containing $\tau$. Note that $\sigma'$ is a rational polyhedral cone, and we are basically in the situation as above, with $\sigma$ replaced by $\sigma'$, except that we don't have any more dimensions to spare, and so $\tau=\sigma'$, as required.