I am having a bit of trouble understanding the proof of Conway induction.
Definition: Descending Game condition: There does not exist an infinite sequence of games $G^i=(L^i,R^i)$ with $G^{i+1}\subset L^i \cup R^i$ for all $i \in \mathbb{N}$.
Theorem: Conway Induction: Let $P$ be a property which games might have, such that any game $G$ has property $P$ whenever all left and right options of $G$ have this property. Then every game has property $P$.
PROOF: Suppose there is a game $G$ which does not satisfy $P$. If all left and right options of $G$ satisfy $P$, then $G$ does also by hypothesis, so there is an option $G_0$ of $G$ which does not satisfy $P$. Continuing this argument inductively, we obtain a sequence $G,G_0,G_1, . . .$ of games, each an option of its predecessor, which violates the Descending Game Condition. Note that formalizing this argument needs the Axiom of Choice.
ISSUE: Why does the sequence have to be infinite. If we reach a game with empty options won't the sequence be finite. Say we start with the game $2\equiv \{1|\}$ then $1\equiv \{0|\}$ then $0$ and the sequence ends.
"whenever all left and right options of $G$ have this property" is vacuously satisfied if $G$ has no left or right options. Therefore by the hypothesis, $0$ has to satisfy the property, and so can never be chosen for this descending sequence.