Let $|X_n|$ < Y a.s., each n, n = 1,2,3..
How could I prove that also $sup_n|X_n|\le Y$ a.s?
How does the proof change if I now that $lim_{n \to \infty} X_n = X$ a.s. and that |X| $ < \infty$? specifically, what I want to prove is now that $sup_n|X_n|<\infty$
I believe you mean $|X_n|\leq Y$ almost surely, not $X_n\leq Y$. (otherwise the statement is obviously false. You need either absolute value at both places or in none of them)
Well, for each $n\in\mathbb{N}$ we have $P(|X_n|\leq Y)=1$. A countable intersection of events with probability $1$ is also an event with probability $1$, hence:
$1=P(\cap_{n=1}^\infty\{ |X_n|\leq Y\})=P(|X_n|\leq Y\ \forall n\in\mathbb{N})=P(\sup_{n\in\mathbb{N}} |X_n|\leq Y)$