Since $(a_n)$ converges, $(|a_n|)$ converges, and thus the limit of the sequence $|a_n|$ is defined. Let a = $\lim_{n \to \infty} |a_n|$. Then by definition of limit, for every $\epsilon > 0$ there exists $N \in \mathbb{N}$ such that $n > N \implies |a_n - a| < \epsilon$ from which it follows that
(1)
$$
-\epsilon < a - a_n < \epsilon \; \forall \; n > N
$$
Since $(a_n)$ is bounded
(2)
$$
-\alpha \leq a_n \leq \alpha \; \forall \; n \in \mathbb{N}
$$
Combining (1) and (2),
$$
|a| < \alpha + \epsilon \; \forall \; \epsilon > 0
$$
Since this equation holds for all positive $\epsilon$ it must be the case that $|a| \leq \alpha$. Moroever, the limit of a nonnegative sequence must also be nonnegative and therefore $|a| = a$ and the claim is proved.