If $\mathcal{F}_1 \subset \mathcal{F}_2 \subset \dotsb$ are sigma algebras, what is wrong with claiming that $\cup_i\mathcal{F}_i$ is a sigma algebra?
It seems closed under complement since for all $x$ in the union, $x$ has to belong to some $\mathcal{F}_i$, and so must its complement.
It seems closed under countable union, since for any countable unions of $x_i$ within it, each of the $x_i$ must be in some $\mathcal{F}_j$, and so we can stop the sequence at any point and take the highest $j$ and we know that all the $x_i$'s up to that point are in $\mathcal{F}_j$, and thus so must be their union. There must be some counterexample, but I don't see it.