Suppose $X_n$ is a sequence of random variables such that $X_n\stackrel{p}{\longrightarrow}X$ with $\mathbb{P}\left[X>0\right]=1$. I want to prove that, for all $\varepsilon>0$, it holds $\mathbb{P}\left[|X_n|\geq \varepsilon\right]\rightarrow 1$.
My attempt: $\varepsilon\leq |X_n|\leq |X_n-X|+|X|$ implies $[\varepsilon\leq |X_n|]\subseteq [\varepsilon\leq |X_n-X|+|X|]$ or
$$\mathbb{P}\left[|X_n|\geq \varepsilon\right]\leq \mathbb{P}[ |X_n-X|+|X|\geq \varepsilon]$$
but this goes in the order direction, since I have to find out a converging-to-one lower bound for $\mathbb{P}\left[|X_n|\geq \varepsilon\right]$.
Any suggestions?
Let $X$ be such that $P(X=1/k)=2^{-k}, k\geqslant 1$ and $X_n=X$. Then for each positive $\varepsilon$, $$ \lim_{n\rightarrow+\infty}P[|X_n|\geqslant \varepsilon]=\sum_{k\geqslant 1: 1/k>\varepsilon}2^{-k}<1. $$ Conclusion: it is possible that $\lim_{n\rightarrow+\infty}P[|X_n|\geqslant \varepsilon]=1$ fail for each positive $\varepsilon$.