- Is the set of all open and closed sets in $\mathbb{R}^2$ a $\sigma$-algebra for $\mathbb{R}^2$?
- Is the set of all closed sets in $\mathbb{R}^2$ a topology for $\mathbb{R}^2$?
I would say no in both cases, since the union of countably many closed sets is not necessarily closed or open again.
You are correct. Let's consider a concrete example. Let $\mathcal{B}_1$ denote the set of all open and closed sets of $\mathbb{R}^2$, and let $\mathcal{B}_2$ denote the set of all closed sets of $\mathbb{R}^2$, where the notions of "open" and "closed" are taken with respect to the usual (metric) topology on $\mathbb{R}^2$.
We use the fact that $A = [0, 1) \times \mathbb{R}$ is neither open nor closed in $\mathbb{R}^2$. $A$ is not open because, for example, there is no neighborhood of $0 \times 0 \in A$ that is contained in $A$ (any neighborhood of $0 \times 0$ also contains, for some $\varepsilon > 0$, the point $-\varepsilon \times 0 \notin A$). $A$ is not closed because its complement $\mathbb{R}^2 - A = ((-\infty, 0) \cup [1, \infty)) \times \mathbb{R}$ is not open (consider the point $1 \times 0$ for example).
Now, we see that $A$ can be written as a countable union of closed sets:
$$A = \bigcup_{n \in \mathbb{Z}^+} \left[0, 1 - \frac{1}{n + 1} \right] \times \mathbb{R}$$
I'll leave it to you to prove the equality. The fact that $A$, a countable union of elements of $\mathcal{B}_1$, is not contained in $\mathcal{B}_1$ shows that $(\mathbb{R}^2, \mathcal{B}_1)$ is not a $\sigma$-algebra. The fact that $A$, an arbitrary union of elements of $\mathcal{B}_2$, is not contained in $\mathcal{B}_2$ shows that $(\mathbb{R}^2, \mathcal{B}_2)$ is not a topology.