Here's the question that I'm trying to answer.
Let $X$ be a plane. Then, let $\Sigma$ be a subset of $P(X)$ such that $X \in \Sigma$, $\varnothing \in \Sigma$ and all open disks centered on the origin are also in $\Sigma$. Is $\Sigma$ a topology on $X$?
So, I think that this is a topology on X. Let me try to prove it by checking the topology axioms:
Clearly, $X \in \Sigma$ and $\varnothing \in \Sigma$. That satisfies the first axiom.
Now, consider any finite intersection of sets in $\Sigma$. We know that $X \cap X = X$ and $X \cap \varnothing = \varnothing$. We also know that $\varnothing \cap \varnothing = \varnothing$. Any intersection of open disks in $\Sigma$ will produce the open disk with the smallest radius in that intersection, so that will still belong to $\Sigma$. Finally, the intersection of any open disk with $X$ will produce the open disk and the intersection of any open disk with $\varnothing$ will produce $\varnothing$. So, $\Sigma$ is closed with respect to the intersection of sets.
Finally, we consider any arbitrary union of sets in $\Sigma$. Any arbitrary union consisting of open disks will produce the largest open disk. Furthermore, any arbitrary union that consists of $X$ and $\varnothing$ will just produce $X$. So, $\Sigma$ is closed with respect to unions of sets.
Since $\Sigma$ clearly satisfies all of the axioms of topological structure, it follows that $(X,\Sigma)$ is a topological space.
Does my proof above work? If it doesn't, how can I improve it so that the argument works? Any feedback would be appreciated.