let's say
$\mathbf{A}=${all intervals contained in [0,1]} and
$\mathbf{B}$={all finite unions of elements of $\mathbf{A}$}
proving that $\mathbf{B}$ is an algebra is mostly straightforward, except i'm not sure how to prove that $\emptyset \subset \mathbf{B}$. And for that matter, how do I even prove that $\emptyset \subset \mathbf{A}$? Can I just say the 'interval' for example $[\frac{1}{2},\frac{1}{2}]$ it not actually an interval is therefore $\emptyset$ ?
To prove that $\mathbf{B}$ is not a sigma-algebra, do I need to construct a countable union or intersection of subsets of $\mathbf{B}$ and show that it's not contained in $\mathbf{B}$? Can I say that $\bigcup_{j \in \mathbf{N}} [\frac{1}{2j+1},\frac{1}{2j}] \notin \mathbf{B}$ (where $\mathbf{N}$ is the set of natural numbers) and thus $\mathbf{B}$ is not closed under countable union and therefore cannot be a sigma-algebra?