$\textbf{The Problem:}$ Show that if a random variable $X$ is such that $P(X=0)<1,$ then there is an $\varepsilon>0$ with $P(\vert X\vert>\varepsilon)>0.$
$\textbf{My Proof:}$ Since $P(X=0)<1,$ we have that $0<P(\{X=0\}^\complement)=P(\vert X\vert>0)\leq1$. Now suppose that no such $\varepsilon>0$ exists. Then for all $n\in\mathbb N$ we have $P\left(\vert X\vert>\frac{1}{n}\right)=0$. Now the sequence $\left(\left\{\vert X\vert>\frac{1}{n}\right\}\right)_{n\in\mathbb N}$ is nested with $\left\{\vert X\vert>\frac{1}{n}\right\}\nearrow\{\vert X\vert>0\}.$ To see this, note that for $n\in\mathbb N$ we have that if $\vert X(\omega)\vert>\frac{1}{n}$, then $\vert X(\omega)\vert>\frac{1}{n+1}$, so $\left\{\vert X\vert>\frac{1}{n}\right\}\subseteq\left\{\vert X\vert>\frac{1}{n+1}\right\}$. Now let $\omega\in\bigcup_n\left\{\vert X\vert>\frac{1}{n}\right\}.$ Then there is $n\in\mathbb N$ such that $\vert X(\omega)\vert>\frac{1}{n}$, and hence $\vert X(\omega)\vert>0$. Thus, $\bigcup_n\left\{\vert X\vert>\frac{1}{n}\right\}\subseteq\{\vert X\vert>0\}.$ For the reverse inclusion let $\omega\in\{\vert X\vert>0\}.$ Then by the Archimedean property there exists $n\in\mathbb N$ such that $\vert X(\omega)\vert>\frac{1}{n}$, so $\omega\in\bigcup_n\left\{\vert X\vert>\frac{1}{n}\right\}.$ Hence, $\bigcup_n\left\{\vert X\vert>\frac{1}{n}\right\}=\{\vert X\vert>0\}.$
Thus, the continuity of the probability measure implies that
$$\lim\limits_{n\to\infty}P\left(\left\{\vert X\vert>\frac{1}{n}\right\}\right)=P(\vert X\vert>0)=0,$$
which is a contradiction. The result follows.
Do you agree with the proof above? If not, it would be much appreciated if anyone could please point out where the proof goes wrong.
Thank you for your time.