How to show a sequence of independent random variables do not almost surely converge by definition? I have a sequence of independent random variables $X_1, X_2, \ldots$ where
$$ X_n =
  \begin{cases}
    1      & \quad \text{with probability} \  1/n  \\
    0    & \quad \text{with probability} \ 1-1/n
  \end{cases}
$$
In my book, it claims that while $X_n \to 0$ in probability, it is not true almost surely. The book gives a very short outline of the proof, but I am not sure if it is correct. It writes that: for $0 < \epsilon <1$, $P\left(\bigcup_{m=n}^\infty \{|X_m - X| > \delta\}\right) = 1- \lim_{n \to \infty}P(X_m = 0 \ \forall m \text{ s.t. } n \leq m)$. I am not sure why this line can show that almost sure convergence to $0$ is not possible. 
The definition of almost sure converge I take is to be: $\forall \epsilon, \delta >0, \exists n_0=n_0(\epsilon, \delta) \text{ s.t. } P(\bigcup_{m=n}^\infty \{|X_m -X|> \delta\})\leq \epsilon$. Can someone tell me what the author's line intends to show? Thank you

That a sequence of members of $\{0,1\}$ approaches $0$ means that after some finite number of terms it is always $0$.
For $\{X_n\}_{n=1}^\infty$ to converge almost surely to $0$, it would have to be the case that
$$
P( \text{for some finite number } N, \text{ for all } n\ge N,\  X_n=0) = 1.
$$
The second Borel–Cantelli lemma tells us that that happens only if $\displaystyle\sum_{n=1}^\infty \frac 1 n<\infty$, but we know that that is false.
