Convergence in distribution equivalent definition

I'm stuck with this exercise.

If $X,X_n$ for $n=1,2,\ldots,$ r.v.such that $X_n \Rightarrow X$. Prove that for all $\delta>0$, exists $M>0$ and $N \in \mathbb{N}$ such that

$$P[|X_n| \geq M]<\delta, \hspace{1cm} \forall n \leq N.$$

I think is straightforward from the definition of limit of a sequence, but I cannot came up with anything as that. Any idea?

I suppose $|X|<\infty$ a.s. As the c.d.f. $F_{|X|}(t):=P(|X|\le t)$ of $|X|$ tends to $1$ as $t\to\infty$, we can take $M>0$ large enough so that $F_{|X|}(M)\ge1-\frac\delta2$, that is $$P(|X|>M)\le\frac\delta2.$$ Moreover, since the set of discontinuity points of $F_{|X|}$ is at most countable, we can assume that $F_X$ is continuous at $M$.

Now the convergence of distribution $X_n\Longrightarrow X$ implies that of $|X_n|\Longrightarrow|X|$ (by the continuous mapping theorem), so $F_{|X_n|}(M)\to F_{|X|}(M)$ as $n\to\infty$. In particular, for all $n\ge N$ sufficiently large, $$F_{|X_n|}(M)\ge F_{|X|}(M)-\frac\delta2.$$ Thus $$P(|X_n|>M)\le\delta.$$