We claim that $\{x + y\} - \{y\}$ is equal to $\{x\}$ or $\{x\} - 1$, for any real numbers $x$ and $y$. Write $x + y = [y] + [x] + \{x\} + \{y\}$ to get $\{x + y\} - \{y\} = \{\{x\} + \{y\}\} - \{y\}$. Hence if $\{x\} + \{y\} < 1$ then $\{x + y\} - \{y\} = \{x\}$, while if $2 > \{x\} + \{y\} \ge 1$ then $\{x + y\} - \{y\} = \{x\} - 1$, as claimed.

Apply the above result to infer that either $|a_{n+1} - a_n| = \{\alpha\}$ or $|a_{n+1} - a_n| = 1 - \{\alpha\}$. Set $b = \max\{\{\alpha\}, 1 - \{\alpha\}\}$ and notice that $0 < b < 1$ (for $\alpha$ is irrational) to derive that $|x_n| \le b^n$, implying $\lim_{n \to \infty} |x_n| = 0$ and furthermore $\lim_{n \to \infty} x_n = 0$.