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 either there exists a hyperplane $H$ containing $\tau$ such that $H$ is not a supporting hyperplane for $\sigma$, or $\tau$ is full-dimensional. If $H$ exists 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). If $H$ does not exist, there exists $x\in\sigma\setminus\tau$ and $y\in\tau$ (again, it's possible to choose them in $N$) so that $x+y\in\tau$, again contradicting (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 either $\tau=\sigma'$, as required, or we contradict (iii) as above.

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 either there exists a hyperplane $H$ containing $\tau$ such that $H$ is not a supporting hyperplane for $\sigma$, or $\tau$ is full-dimensional. If $H$ exists 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). 

If $H$ does not exist, then $\dim \tau=\dim\sigma$, and there exists $x\in\sigma\setminus\tau$ ($x$ may be chosen to be a generator for an extreme ray of $\sigma$) and $y\in\tau\setminus\partial\tau$ (again, it's possible to choose $x,y\in N$) so that $x+y\in\tau$, again contradicting (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 either $\tau=\sigma'$, as required, or we contradict (iii) as above.

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.

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 either there exists a hyperplane $H$ containing $\tau$ such that $H$ is not a supporting hyperplane for $\sigma$, or $\tau$ is full-dimensional. If $H$ exists 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). If $H$ does not exist, there exists $x\in\sigma\setminus\tau$ and $y\in\tau$ (again, it's possible to choose them in $N$) so that $x+y\in\tau$, again contradicting (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 either $\tau=\sigma'$, as required, or we contradict (iii) as above.