Let $\{X_n\}_{n \in \mathbb{N}}$ be a sequence of random variables that converges almost surely to the variable X. Prove that $X_n$ converges to $X$ in probability.
My attempt
Let $\varepsilon >0$. For each $n \in \mathbb{N}$ let us define the events $$A_n = \bigcup_{k\geq n}(|X_k - X| \geq \varepsilon)$$
This sequence is such that $A_{n+1} \subset A_n$ for all $n \in \mathbb{N}$. Since $(|X_n - X|\geq \varepsilon ) \subset A_n$, then $P(|X_n - X|\geq \varepsilon ) \leq P(A_n)$, where $P$ is the probability measure of the sets. Therefore, \begin{align*} \lim_{n \to \infty}P(|X_n - X|\geq \varepsilon ) &\leq \lim_{n \to \infty}P(A_n)\\ &=P\left(\lim_{n \to \infty}A_n\right)\\ &=P\left(\bigcap_{n \geq 1}A_n\right)\\ &= P(|X_n -X|\geq \varepsilon, \forall n\geq 1)\\ &= P\left(\lim_{n \to \infty}X_n \neq X\right)\\ &= 0 \end{align*}
My question is this: why $P(\bigcap_{n\geq 1}A_n) = P(|X_n - X| \geq \varepsilon, \forall n\geq 1)$
That is, how to justify this step of the proof? Or, what is the need to do this step?
In order for me to prove that these probabilities are equal, I think I should prove that $\bigcap_{n\geq 1}A_n = \{ |X_n - X| \geq \varepsilon, \forall n\geq 1\}$, however this is not true in general as can be seen in the comments. How can I do that? I would be very grateful if someone can help me understand this part.
The intermediate step in your derivation is plain false. The correct one is the following. $$ \bigcap_{n\geq 1} A_n = \bigcap_{n\geq 1} \bigcup_{k\geq n}\{|X_k - X| \geq \varepsilon\} = \{\forall n\geq 1, \exists k\geq n, |X_k - X| \geq \varepsilon\} = \{X_n \not \to X\} $$