Here's a proof assuming $x$ is rational. Write $x=\frac{a}{b}$ for $a,b\in\mathbb{N}$ in lowest terms. Choose $\epsilon>0$ such that $x\epsilon<\frac{1}{2b}$ and $\epsilon<\frac{1}{2b}$ (actually, the latter is automatic since $x\geq 1$). Choose $N$ such that $|\alpha x^n-m_n|<\epsilon$ for all $n\geq N$ and such that $m_N\neq 0$ (the latter is possible since $x\geq 1$ and $\alpha\neq 0$). In particular, for any $n\geq N$, we can write $$\alpha x^n=m_n+\delta$$ where $|\delta|<\epsilon$ ($\delta$ depends on $n$). Multiplying by $x$, we get $$\alpha x^{n+1}=xm_n+\delta x.$$ Note that $xm_n$ is a rational number with denominator $b$, and $|\delta x|<\frac{1}{2b}$. If $xm_n$ were not an integer, then it would be at least $\frac{1}{b}$ away from an integer, and so $xm_n+\delta x$ would be at least $\frac{1}{2b}$ away from an integer. But by hypothesis, $\alpha x^{n+1}$ is within $\epsilon$ of $m_{n+1}$, and $\epsilon<\frac{1}{2b}$. Thus $xm_n$ must be an integer, and clearly in fact we must have $xm_n=m_{n+1}$.
Thus $xm_n=m_{n+1}$ for all $n\geq N$. But this means $x^km_N=m_{N+k}$ is an integer for all $k\in\mathbb{N}$. This implies $m_N$ is divisible by $b^k$ for all $k$. Since $m_N\neq 0$, this is only possible if $b=1$, i.e. if $x$ is an integer.
