As indicated in the comments, you have to state some additional assumptions for this to be interesting. This answer assumes $\alpha_n \neq 0$ for all $n$ and $P(\vert X_n\vert < \infty) = 1$ for all $n$.

Fix $n$ and observe / prove that for every $\epsilon_n >0 $ there exists a constant $M_n = M_n(\epsilon_n)$ such that $P(\vert X_n \vert > M_n) \leq \epsilon_n$. That is,

$$ P(\vert \alpha_n X_n\vert>\vert \alpha_n\vert M_n)\leq \epsilon_n. $$

Can you pick $\epsilon_n$ and $\alpha_n$ so as to make $P(\vert \alpha_nX_n\vert > \delta) \to 0$ for any fixed $\delta>0$?

As indicated in the comments, you have to state some additional assumptions for this to be interesting. This answer assumes $\alpha_n \neq 0$ for all $n$ and $P(\vert X_n\vert < \infty) = 1$ for all $n$.

Fix $n$ and observe / prove that for every $\epsilon_n >0 $ there exists a constant $M_n = M_n(\epsilon_n)$ such that $P(\vert X_n \vert > M_n) \leq \epsilon_n$. That is,

$$ P(\vert \alpha_n X_n\vert>\vert \alpha_n\vert M_n)\leq \epsilon_n. $$

Can you pick $\epsilon_n$ and $\alpha_n$ so as to make $P(\vert \alpha_nX_n\vert > \delta) \to 0$ for any given $\delta>0$?