Let $D_\infty$ be generated by $a$ the reflection fixing the line $\{x=0\}$ and $b$ the reflection fixing the line $\{x=1\}$ in $\mathbb{R}^2$. I want to show that any $\alpha\in \text{Aut}(D_\infty)$ maps $a$ and $b$ to adjacent reflections, so $\alpha(a)$ is the reflection fixing $\{x=n\}$ and $\alpha(b)$ is the reflection fixing $\{x=n\pm 1\}$ for some $n\in\mathbb{Z}$.
In general, the reflection fixing $\{x=n\}$ is $(ba)^{n-1}b$ and the reflection fixing $\{x=-n\}$ is $(ab)^na$, $n\in\mathbb{N}$.
If $\alpha\in\text{Aut}(D_\infty)$, then $\alpha(a)^2=\alpha(a^2)=\alpha(e)=e$, where $e$ is the identity, proving that $\alpha(a)$ and $\alpha(b)$ are reflections again. I'm having trouble proving that they are adjacent. Going case by case, if $\alpha(a)=(ba)^nb$, $\alpha(b)=(ba)^m b$, then $\alpha(ab)=\alpha(a)\alpha(b)=(ba)^n(ab)^m=(ab)^{m-n}$. If $\alpha(a)=(ba)^nb$, $\alpha(b)=(ab)^ma$, then $\alpha(ab)=(ba)^{n+m+1}$. The other two cases are similar, so $\alpha(ab)$ always has the length $\pm2(m-n)$ or $2(n+m+1)$.
Maybe I am overcomplicating this, but I would like to argue with the length of the expressions for $ab$ now. The length of the expression for $ab=(\alpha^{-1}(ab))^{m-n}$ must be the product of $2$ and ($\pm(k-l)$ or $(k+l+1)$) and ($\pm(n-m)$ or $(n+m+1)$), but this product has to be equal to $2$, so that the only possibility is that $n=m+1$ or $m=n+1$ or $n=m=0$.
Would this be a valid argument?