I'm trying to prove that convergence almost surely implies convergence in probability. I know this proof can be found everywhere online, but I don't find one using the fact that:
$$ X_n \rightarrow_{n+\infty}^{p.s.} X \Rightarrow \forall \epsilon > 0, \lim_{n+\infty} \mathbb{P}(\sup_{k\geq n} \lvert X_k - X \rvert \geq \epsilon) = 0 $$
My attempt:
Let's fix $\epsilon > 0$. Let $n \in \mathbb{N}$.
$$\lvert X_n - X \rvert \leq \sup_{n} \lvert X_n - X \rvert$$
Hence,
$$ (\sup_n \lvert X_n - X \rvert < \epsilon) \subset (\lvert X_n - X \rvert < \epsilon)$$
Using the fact that the probability mapping is monotonically increasing,
$$\mathbb{P}(\sup_n \lvert X_n - X \rvert < \epsilon) < \mathbb{P}(\lvert X_n - X \rvert < \epsilon)$$
yet,
$$\lim_{n+\infty} \mathbb{P}(\sup_{k\geq n} \lvert X_k - X \rvert < \epsilon) = 1$$
So,
$$\lim_{n+\infty} \mathbb{P}(\lvert X_n - X \rvert < \epsilon) = 1$$
Eventually,
$$\lim_{n+\infty} \mathbb{P}(\lvert X_n - X \rvert \geq \epsilon) = 0$$
So, $(X_n)$ converges in probability towards $X$.
I don't know if it is correct, I feel like I don't manage well the supremum.