Error in solution of Peter Winkler "red and blue dice" puzzle?

The figure is mistaken, but the proof is not, after a small clarification. Ignore the figure.

Winkler intended to assume $\alpha_n\le \beta_n$. Furthermore, he intended $0$ to be an allowable index when choosing $m'$, where $\beta_0=0$. This ensures ${m'}$ exists. For each $m\ge 1$, we have $\alpha_m \ge \beta_0$. This implies that $\{i:0\le i\le n,\alpha_m\ge \beta_i\}$ is nonempty; possibly it only contains $i=0$. We then let $m'$ be the largest element of this set.

By definition, $\alpha_m-\beta_{m'}\ge 0$. If $\alpha_m-\beta_{m'}\ge n$, then it must be that $m'<n$, because $\alpha_m\le \alpha_n \le \beta_n$. We can then consider $\beta_{m'+1}$, and would have $\beta_{m'+1}=\beta_{m'}+((m'+1)^{st}\text{ dice})\le \beta_{m'}+n\le \alpha_m$, contradicting the maximality of $m'$. Therefore you have $0\le \alpha_m-\beta_{m'}\le n-1$ and the rest of the proof follows.

It seems to me that there are two holes in this proof. The first is in the statement that the labels must be less than $n$. Suppose that $\alpha_n-\beta_n\ge n.$ Then $n'=n,$ and there is no larger index available. Then perhaps $\alpha_n\le\beta_n$ is right after all, and the diagram is wrong.

Yes, take $\alpha_n\leq \beta_n.$

But this leaves the second problem, which doesn't depend on the relation between $\alpha_n$ and $\beta_n.$ Suppose that $a_1<b_1.$ How is $1'$ to be defined?

Set $1'=0.$ When you take the intervening dice between two different "$m$"'s the lower "$m$" is excluded, so it's fine to use zero here. The higher "$m$" is included, but that's ok: if $\alpha_m-\beta_{m'}=\alpha_M-\beta_{M'}=c$ with $m<M$ then necessarily $M'>0$ because $\beta_{M'}>\beta_{m'}.$