Help determining if a finite subset of $\mathbb R$ is closed and bounded. 
If $\{A_n \; : \; n \in \mathbb N\}$ is any collection of subsets of $\mathbb R$, with each set $A_n$ containing finitely many numbers, then the union $\bigcup_{n=1}^{\infty}A_n$ is closed and bounded.

I need to determine if this is true or false and if true prove it, if false provide a counter-example.  
I believe this is a true statement because every finite subset of $\mathbb R$ is a bounded set, but I'm not sure if it is a closed set?  I imagine it too is closed because every limit point would be in $S$.
Assuming this is true, how would I start off proving it?

If $(r_n)$ is any Cauchy sequence of real numbers and $(s_n)$ is any increasing sequence of negative numbers, then the sequence $(r_n-s_n)$ converges.  

I'm confused by this one because is $s_n$ "increasing" negatively or positively?  In other words if I'm at $-1000$ is my sequence heading towards $0$ or $-100000$?  What does increasing mean in this context?
Since it requires a Cauchy sequence of $\mathbb R$ a good example would would be $\frac{1}{n}$
 A: If $A_n = \{ n \}$ then $\bigcup_{n=1}^{\infty} A_n = \mathbb{N}$ which is not bounded so the union need to be bounded but it is closed in this case.
If $A_n = \{ \frac{1}{n} \}$ then $\bigcup_{n=1}^{\infty} A_n$ is bounded but not closed ($0$ is a limit point). 
It is a good exercise to construct an example of $A_n$ in which $\bigcup_{n=1}^{\infty} A_n$ is not closed and also not bounded at the same time.
A: 1) Hint:  Any countable subset $V \subset \mathbb R$ can be written as $V = \cup_{v_i \in V} \{v_i\}$ any countable set can be such a union.
Are all countable subsets of $\mathbb R$ closed and bounded? Can you prove it? Can you find a counter example?
(Your reasoning that if every finite set is something then a countable union of finite sets is something is just plain wrong.  An obvious example is every finite set is finite but a countable union of finite sets is not.  Bounded and finite are very similar concepts.)
2) Increasing means $a_{i+1} > a_i$.  As $-999 > -10000$ we are going toward the zero.  (The statement is obviously false otherwise).
So the sequence $\{s_n\}$ is increasing yet also negative.  What does that tell you?  It's heading toward zero.  Does it ever get to zero?  Does it ever get above zero?  What's the terminology for that?  What does that imply about the sequence $\{s_n\}$?
