Find $n$ such that $\{n\alpha\}=n\alpha-\lfloor n\alpha\rfloor<\frac 1m$.
Then $0
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. $-12$ 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 $k>\frac{1+a-n_0\alpha}{\alpha-\frac d{n_1-n_0}}$.