Is my attempt correct? Let $\xi_{1}, \xi_{2}, \ldots$ be a sequence of random variables on a probability space
$(\Omega, \mathcal{F}, \mathrm{P})$ such that $\mathrm{E}\left|\xi_{n}\right| \leq 2^{-n} .$ Prove that $\xi_{n} \rightarrow 0$ almost surely as
$n \rightarrow \infty .$
my attempt : $E |\xi_{n}| \leq 2^{-n}$, therefore by the dominated convergence theorem we have : 
$\int_{\Omega} \lim_n |\xi_n| d \mathcal{P} = 0 $ 
this implies that : $\lim_n |\xi_n| = 0 \text{ a.s. }$
so $\xi_{n} \rightarrow 0$ almost surely as
$n \rightarrow \infty .$

You have a flaw in your argument,because you assume that the limit of $|\xi_n|$ exists.
Here is what you can do:
By monotone convergence,
$$E(\sum_n|\xi_n|)=\sum_n E(|\xi_n|)<+\infty$$
So $\sum_n|\xi_n|<+\infty$ a.s 
Thus $\xi_n \to 0$ a.e
