Since $X_n$ and $X$ are both measurable (they are random variables after all) and the function $f(x,y) = \lvert x-y \rvert$ preserves measurability (due to its continuity), $\lvert X_n - X \rvert$ is measurable as well.
Maybe it is better for your understanding if I argue in the following way. If $X_n$ and $X$ are measurable, then so is $X_n - X$. This is a standard result in measure theory. Then, $\{ \lvert X_n - X \rvert > \varepsilon \} = \{ X_n - X > \varepsilon \} \cup \{ X_n - X < -\varepsilon \}$. Both elements on the right hand side lie in the $\sigma$-algebra defined as part of the probability space in which $X_n$ and $X$ live. Let's call this $\sigma$-algebra $\mathcal{F}$. You know from the definition of a $\sigma$-algebra that the countable union keeps you inside the $\sigma$-algebra. So then the left hand side must be in $\mathcal{F}$, i.e. it is an event.