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For each history of the first two votes, voter 3 selects its favourite achievable winner. Writing rows for voter 1's vote and columns for voter 2's vote, the resulting continuation winners areFor example, after votes $B,C$, voter 3 can make either $B$ or $C$ win, and prefers $B$; a vote for $A$ would instead produce the tie-break winner $A$.

Voter 2 prefers $C$ to $B$ to $A$. Optimizing each row therefore gives continuation winners $C,B,C$ after voter 1's respective choices $A,B,C$. Voter 1 prefers $B$ to $C$ and therefore votes $B$, giving winner $B$ in every subgame perfect equilibrium.

Some histories admit several equally good votes, but they produce the same winning candidate. Since each voter strictly ranks the three possible winners, unique outcome under strict outcome preferences makes this winner independent of those action ties. This is backward induction over outcomes, not a claim that the entire equilibrium strategy profile is unique.

Yes: let the order be voter 2, then voter 3, then voter 1. When voter 1 moves last, it makes $A$ win whenever that is possible; otherwise it prefers $B$ to $C$. For the first two votes in this new order, its optimal continuation winners areRows now record voter 2's initial choice and columns voter 3's choice. In a three-way split, the specified tie-break produces $A$, which explains, for example, the off-diagonal $B,C$ entry.

Voter 3 optimizes its ranking $B\succ C\succ A$ in each row, giving the continuation winners $A,B,C$. Voter 2 then chooses $C$, its favourite of the three. The winner is therefore $C$ in every subgame perfect equilibrium, again by unique outcome under strict outcome preferences. In particular, if voter 2 starts with $C$, voter 3 must also vote $C$: either other vote lets the last voter force $A$.