Find $n$ such that $\{n\alpha\}=n\alpha-\lfloor n\alpha\rfloor<\frac 1m$. Then $0 for $k=1,\ldots, m$, hence $\lfloor kn\alpha\rfloor =k\lfloor n\alpha\rfloor$.
Assume there is an infinite arithmetic progression, i.e. we have $a,d\in\mathbb Z$, $d\ne 0$ such that $a+k d\in A$ for all $k\in\mathbb N_0$. Select $n_k$ such that $\lfloor n_k\alpha\rfloor=a+kd$. Note that $\alpha>1$ implies that $n_k$ is uniquely determined. Let $m_k=n_0+k(n_1-n_0)$.
Claim: $m_k=n_k$. This is clear for $k=0$ and for $k=1$. Assume we know already $m_i=n_i$ for all $i\le k$. Then $\begin{align}m_{k+1}\alpha &= (m_k+n_1-n_0)\alpha\\ &=n_k\alpha+n_1\alpha-n_0\alpha\\ &=(a+kd)+\{n_k\alpha\}+(a+d)+\{n_1\alpha\}-a-\{n_0\alpha\}\\ &=a+(k+1)d+\{n_k\alpha\}+\{n_1\alpha\}-\{n_0\alpha\},\end{align}$ i.e. $-1. Since also $0, we conclude $-2<(m_{k+1}-n_{k+1})\alpha<2$ and with $\alpha>2$ we obtain $m_{k+1}=n_{k+1}$. This proves the claim.
Thus we find that $0<(n_0+k(n_1-n_0))\alpha-(a+kd)<1$ for all $k\in\mathbb N$. But this implies $0<\alpha-\frac d{n_1-n_0}<\frac{1+a-n_0\alpha}k$ which is absurd if we choose $k>\frac{1+a-n_0\alpha}{\alpha-\frac d{n_1-n_0}}$.