Is uncountable union of $\sigma$-algebra necessarily algebra? It is well known that infinite union of $\sigma$-algebra is not necessarily a $\sigma$-algebra. 
But then I read this:

The countable union of a non-decreasing sequence of σ-algebras is an algebra.

I am not sure why countable union is needed here.
Is there any example that uncountable union of $\sigma$-algebra that is not an algebra?
 A: Countability is totally irrelevant here, but what is essential is that the algebras form a "non-decreasing sequence".  What this should mean for more general (not necessarily countable) collections is that the algebras you are taking the union of are totally ordered under inclusion.  More precisely, we have the following:

Let $X$ be a set and let $\mathcal{C}$ be any nonempty set of algebras on $X$ such that for any $A,B\in\mathcal{C}$, either $A\subseteq B$ or $B\subseteq A$.  Then $\bigcup\mathcal{C}$ is also an algebra.

To prove this, first note that since $\mathcal{C}$ is nonempty, $\emptyset$ and $X$ are in $\bigcup\mathcal{C}$.  Now suppose $S,T\in\bigcup\mathcal{C}$.  Then for some $A,B\in\mathcal{C}$, $S\in A$ and $T\in B$.  Without loss of generality, $A\subseteq B$.  Then $S,T\in B$, so $S\cup T, S\cap T$, and $X\setminus S$ are all in $B$.  Thus they are all in $\bigcup\mathcal{C}$, which says that $\bigcup\mathcal{C}$ is an algebra.
If you remove the assumption that $\mathcal{C}$ is totally ordered, then this result can fail already for finite unions, even if the elements of $\mathcal{C}$ are $\sigma$-algebras.  For instance, take $X=\{0,1,2\}$ and $\mathcal{C}=\{A,B\}$ where $A=\{\emptyset, \{0\},\{1,2\},X\}$ and $B=\{\emptyset,\{1\},\{0,2\},X\}$.
(Actually, you don't quite need $\mathcal{C}$ to be totally ordered; it suffices for it to be directed: for any $A,B\in\mathcal{C}$, there exists $C\in\mathcal{C}$ such that $A,B\subseteq C$.  The proof is similar.)
