A is a plane. B is the set of all open disks. Is B a topology on A? Let A be a plane. Let B contain all open disks centered at the origin of the plane. B also contains the empty set and A. Is B a topology?
To make sure B is a topology we have to prove:
The empty set and the whole set A is contained in B. (This we already know)
Any union of elements in B is in B. (So if we take the union of the open disk of radius 2 and radius 10 the union is in B)
Any intersection of finitely many elements in B is in B. (The intersection of the open disk 2 and radius 10, which is the open disk 2 is in B)
What I'm having trouble with is proving that the union of any open disk is in A. I mean we know that any open disk of any size will be on the plane but idk how to write it in a proof. I was thinking of doing something with radii and saying that all the points less than the radii are in the plane thus its contained in B but how can I say this arbitrarily?
 A: Let $D_\alpha$ be a family of open disks centered at the origin, where $\alpha\in \Lambda$ and $\Lambda$ is any set. WTS $\cup_{\alpha\in A}\; D_\alpha\in B $ 
For all $D_\alpha$, let $r_\alpha$ be the radius. Then $(r_\alpha)_{\alpha\in \Lambda}\subseteq \mathbb{R}^+\cup \{0\}$. As such, $sup((r_\alpha)_{\alpha\in \Lambda})$ is well defined, though possibly infinite. Now if the supremum is infinite then your union of the disks is $A$, which you know is in $B$. Notice that since you are taking the supremum over a non-empty set, the supremum will not be $-\infty$. Then the supremum is finite, and you can call it $r$. You have that $x\in\cup_{\alpha\in \Lambda}\; D_\alpha$ implies $x\in D_{\bar{\alpha}}$ for some $\bar{\alpha}$. This means $|x|<r_{\bar{\alpha}}$. As such $|x|<r$, implying $x\in D_r$. Hence $\cup_{\alpha\in \Lambda}\; D_\alpha\subseteq D_r$. Now if $x\in D_r$ then $|x|<r$ and by the definition of supremum $|x|<r_{\bar{\alpha}}$ for some $\bar{\alpha}$. As such, $x\in D_{\bar{\alpha}}$ implpying $x$ is in the onion. Hence $\cup_{\alpha\in \Lambda}\; D_\alpha= D_r$. Therefore, the arbitrary union is an open disk, hence in $B$. 
A: Let $D(x,r)$ denote the disc of radius $r$ centered at a point $x$, and write $0 := (0, 0)$ for the origin of the plane $A := \mathbb{R}^2$.
Suppose we have a nonempty set of open discs centered at the origin of radius $r_\alpha$ for $\alpha \in \Lambda$.  We want to show that $\bigcup_\alpha D(0, r_\alpha)$ is either an open disc centered at the origin or that it is the entire plane.
Write $R := \sup \Lambda$.  Since $\Lambda$ is nonempty this is either a finite number or $+\infty$.  In the latter case I claim that $\bigcup_\alpha D(0, r_\alpha) = \mathbb{R}^2$.  To see this, suppose that $(x, y) \in \mathbb{R}^2$.  Let $r := \sqrt{x^2 + y^2}$.  Since $R = + \infty$, there is some $\alpha \in \Lambda$ with $r_\alpha > r$.  Consequently,
$$(x, y) \in D(0, r_\alpha) \subseteq \bigcup_\alpha D(0, r_\alpha).$$
Since $(x, y)$ was an arbitrary point of $\mathbb{R}^2$, we have $\mathbb{R}^2 \subseteq \bigcup_\alpha D(0, r_\alpha)$, and of course the reverse containment is automatic.
So suppose on the other hand that $R$ is finite.  I claim that $\bigcup_\alpha D(0, r_\alpha) = D(0, R)$.  To see this, suppose $(x, y) \in D(0, R)$. [you fill in the rest]
