To prove that this is a metric, consider the following theorem.

**Theorem.** If the second player can survive for $n$ steps in the game with $\Gamma_1$, $\Gamma_2$, and for $m$ steps with $\Gamma_2$,$\Gamma_3$, then he can survive for $\min(n,m)$ steps with $\Gamma_1$, $\Gamma_3$. 

**Proof.** The idea is simply to combine the strategies for the two games. Fix strategies for the second player in the $(\Gamma_1,\Gamma_2)$ and $(\Gamma_2,\Gamma_3$ games. In the main $(\Gamma_1,\Gamma_3)$ game, now, let the second player answer any move of the opponent on either side by first copying the move into $\Gamma_2$, and then copying the response of that move into $\Gamma_1$ or $\Gamma_3$, accordingly. In this way, every play of the $(\Gamma_1,\Gamma_3)$ game can be seen as the composition of the strategies for $(\Gamma_1,\Gamma_2)$ and $(\Gamma_2,\Gamma_3)$ games. Since the resulting finite subgraphs $A_1$ and $A_2$ will be isomorphic for $n$ steps and $A_2$,$A_3$ isomorphic for $m$ steps, we will thereby maintain $A_1$ isomorphic to $A_3$ for $\min(n,m)$ steps, as desired. $\Box$

It now follows that the distance from $\Gamma_1$ to $\Gamma_3$ is no larger than the shortest distance from $\Gamma_1$ to $\Gamma_2$ or $\Gamma_2$ to $\Gamma_3$, and so we will obey the triangle inequality.