Soft question: Union of infinitely many closed sets this is a question that is not addressed in my book directly but I was curious. We just proved that the union of a finite collection of closed sets is also closed, but I was curious about if the union of infinitely many closed sets can be open. This question may not be at the level of the book so perhaps that's why it wasn't addressed.
Just to make things easier, lets imagine sets that are disks in the x-y plane. I can imagine that if there are nested disks inside each other, that in this case the union would clearly be closed.
But what if you could construct an infinite set of disks that together cover the entire real plane. Then in this case, it seems that every point in their union would have an open ball centered around the point that is also contained in the real plane, so that this union of an infinite collection of disks would create an open set.
Is this a correct way of thinking? Or at least on the right track? 
I get the feeling that as long as you have no largest individual set that contains all the others, then you won't get a closed set. But I have a feeling there is more subtlety to it.
Thanks everyone
 A: 
We just proved that the union of a finite collection of closed sets is also closed, but I was curious about if the union of infinitely many closed sets can be open.

Sure, one of the common examples being $\mathbb{R}$ with the usual Euclidean topology,
$$(0,1) = \bigcup_{n=2}^\infty\left[\frac{1}{n},1-\frac{1}{n}\right]\text{.}$$
A: Let $U$ be any open set whatsoever.  Each point $x \in U$, taken as a one-point set $\{x\}$, is closed, and $U$ is the union of all these sets.
A: In $\Bbb R$ take $\aleph_0$ closed intervals from $-n$ to $n$ for natural $n$: $$\bigcup_{n\in\Bbb N} [-n,n] = \Bbb R$$
Similary $$\bigcup_{n\in\Bbb N^+} \left([-n,-n+1] \cup [n-1,n]\right) = \Bbb R$$
A: Just to address your "feeling" -- using $\mathbb{R}$ as the counter-example might feel like a bit of a trick or a special case, since it's both closed and open. But remember that in $\mathbb{R}$ (and in general in $\mathbb{R}^n$) there are lots of infinite closed sets. Any half-closed interval whose other end is infinity, $[x,\infty)$ or $(-\infty, x]$, is a closed set, and you can easily see how to express those as an infinite union of closed finite sets. So there's no need for any of the things in the union to bound the rest.
Moving on to smaller closed sets, you can express $[0, 1]$ as an infinite union of closed sets, none of which contains all the others. Consider $[1/n, 1]$ for $n > 1$ together with $[0, 0]$. 
Your feeling was (not quite) "every closed cover of a closed set has a sub-cover of size 1". That's false, and putting it that way might not mean much to you right now. But remember it for when you hear the definition of a compact set, because you were almost on to something important.
