Prove For $(X_n)_{n \geq1}$ independent RVs, $X_n \rightarrow X \ \text{ a.s.} \Rightarrow \sum _{n\geq1} P(|X_n -X| \gt \varepsilon) \lt \infty$

I would like to show the following:

For $(X_n)_{n \geq1}$ independent RVs, $$X_n \rightarrow X \ \text{ a.s.} \Rightarrow\ \forall \varepsilon \gt0, \ \sum _{n\geq1} P(|X_n -X| \gt \varepsilon) \lt \infty$$

We don't know that $|X_n -X| \gt \varepsilon$ are independent so second Borel-Cantelli cannot be applied. Any hint is appreciated.

You can show that the events $|X_n-X|>\epsilon$ are independent. This is because you can use Kolmogorov's $0-1$ law to show that $X$ is constant almost surely.
$X$ is measurable with respect to the tail $\sigma$-algebra $$\bigcap_{n\ge 1}\sigma(X_n,X_{n+1},X_{n+2}\dots).$$
(Intuitively, in order to know the limit of $X_n$, it suffices to know the values of all but any finite subset of the $X_n$.)
Kolmogorov's $0-1$ law says that all events in the tail $\sigma$-algebra of series of independent random variables have probability $0$ or $1$. Therefore, for any $c\in \mathbb R$, $P(X\le c)=0\text{ or }1$, so $X$ is constant almost surely.