Let $X$ be any set, and $A \subset X$.
Is $\tau = \{ X , \emptyset , A,A^{c} \}$, a topology on $X$?
And if so, would it also constitute a $\sigma$-algebra consisting of only open sets?
Edit:
Furthermore, if $\tau = \{X,\emptyset,A_{1},A,_{2},\dots\}$ such that $A_{i} \cap A_{j} = \emptyset$ for all $i \neq j$ and $\cup_{n} A_{n} = X$, would the aboves still hold true?
The first topology is indeed a topology, since any intersection of open sets is open and finite unions are again open. It does contain $X$ and $\emptyset$, so it does form a topology.
The second, however, does not necessarily define a topology: as it is stated, finite unions of elements $A_1...A_k$ for some $k$ do not look like they are necessarily open as well. In order to make it a topology you need to define more properties regarding the sets $A_i$.
Yes, the first one is also a $\sigma$-algebra since it is a finite-element topology closed under complement operation; however, for the same reason above, the second may not be.