TITLE: Prove $\mathbb{R}$ is connected
QUESTION [5 upvotes]: Prove that $\mathbb{R}$ is connected. PLease i have found other ways to prove it but i want to make this way work.
Proof:  
1) Strategy : If i show that a arbitrary interval is connected then  i can take the colection of intervals around zero that make up $\mathbb{R}$ and have a common point So that the union is connected.
TO show that $(a,b)$  is connected ill use the fact that if it is not connected then There is an clopen in $(a,b)$ that makes up a separation with its complement in $(a,b)$ and there is not limit point of one another in any of the 2.Then ill use Least upper bound of property of that clopen set and come to a contradiction.
2) Let $(a,b)$ arbitrary open interval in $\mathbb{R}$. Suppose  $\mathbb{R}$ is not connected. And also $(a,b)$ is not connected.Since it is not connected there is a clopen set $V \subseteq (a,b) $ such that it makes a separation  $V \cup V^{c} =(a,b) $ . NOw since $V$ is open in $(a,b)$ $$V=K \cap (a,b)$$ and $$U=C \cap (a,b) $$ where $U=V^{c}$  and $K,C$ open in $\mathbb{R}$.
Now i have $V \cap U= \emptyset $ $$(K \cap (a,b)) \cap  (C \cap (a,b)=\emptyset $$ so $$(K \cap C) \cap (a,b) =\emptyset  $$  
Now for the last to be true and also that my $U$ and $V$ are non empty  and by drawing and trying to figure it out the $ K \cap C$ must be empty otherwise  $V$ or $U$ will be empty. I can see that with drawing intervals but i cant prove it . 
Hence i need to prove that $K \cap C = \emptyset $  .After i know that  the specific intersection is empty  things are much clear and  i can take the supremum of $U$ and it will be  a limit point of $U$ and be in $U$ but it will be also limit point of $V$ since every open area of it will have a non empty intersection. So it cant be a separation of $(a,b)$ and that means the only  clopen sets of $(a,b)$ are itself and the empty hence it is connected and complete the proof.
Is what im trying to prove even a necessary setp? Or i can just go on with the supremum argument straight away.But  i really wanna prove that intersection is empty i draw all the possibilities and it has to be empty.It is so frustrating something so easily seen not being able to write it down??? 
I dont want proofs  using other arguments. I could easily show $R$ is path connected or myabe some other proofs. Im just stuck trying to do this one so you kinda feel  my pain.

REPLY [2 votes]: Proving that $\mathbb{R}$ is connected is no harder than proving connectedness of any open interval.
Suppose $\mathbb{R}$ (or an open interval $(p,q)$ as well) is the disjoint union of two nonempty open sets $A$ and $B$.
Fix $a\in A$ and $b\in B$; it's not restrictive to assume $a<b$.
Consider $c=\sup\{x\in A:x<b\}$. It exists because the set $C=\{x\in A:x<b\}$ is not empty (it contains $a$) and upper bounded by $b$. Note that, by construction, $c\le b$.
There are two cases: either $c\in A$ or $c\in B$.
Suppose $c\in A$. Then $c<b$ and there exists $\delta>0$, $\delta<b-c$, such that $(c-\delta,c+\delta)\subseteq A$. In particular, $c+\delta/2\in A$ and $c+\delta/2<b$: a contradiction to $c=\sup C$.
Suppose $c\in B$. Then there exists $\delta>0$ such that $(c-\delta,c+\delta)\subseteq B$. By definition of supremum, there is $a'\in C$ with $a'>c-\delta$. But $C\subseteq A$ by assumption, so $a'\in A\cap B$, a contradiction.