Proof- Disjointed open sets in $\mathbb{R}$ are separated. Definition: If two sets $A,B \subseteq \mathbb{R}$ are separated if $\bar{A} \cap B = \emptyset$, and ${A} \cap \bar B = \emptyset$.
Statement: If two open sets $U,V \subseteq \mathbb{R}$ are open and disjoint prove they are separated. 
Proof: We need to prove that $\bar{U} \cap V = \emptyset$. 


*

*Let's assume that $\bar{U} \cap V \neq \emptyset$. 

*This implies that a limit point of $U$ say $x$,  lies in V, $x\in V$. 

*Since it is a limit point of $U$ there is a sequence $\{x_n\}\in U:(x_n)\rightarrow x$. 

*Since $x\in V$ it implies there an neighborhood   $\rho_r(x) \in V$, which is completely contained in $V$. 

*Since $\{x_n\}$ converges to $x$ there exist some $N$, such that for all $n > N, {x_n} \in \rho_r(x)$. 

*This would mean that $U \cap V \neq \emptyset$. 


This is a contradiction, and hence our assumption must be incorrect. Hence $\bar{U} \cap V = \emptyset$. And by symmetry $\bar{V} \cap U = \emptyset$. Hence open disjointed sets are separated. 
I would like to know if the proof is complete and if I can make it better.
 A: This proof is basically correct and complete.  Here are some quite minor nitpicks (there are a lot of them, but they really are minor and your proof is pretty much fine).


  
*This implies that a limit point of $U$ say $x$,  lies in V, $x\in V$. 
  

There is no reason to write $x\in V$ at the end here (and doing so is ungrammatical), since you already said $x$ lies in $V$.  Also, a priori you only know that either a point of $U$ or a limit point of $U$ lies in $V$, since $\overline{U}$ contains both the points of $U$ and the limit points.  However, in this case you must have a limit point since $U$ and $V$ are disjoint (and also every point of an open set is a limit point).


  
*Since it is a limit point of $U$ there is a sequence $\{x_n\}\in U:(x_n)\rightarrow x$. 
  

Your notation seems a bit off here: is your notation for sequences $\{x_n\}$ or $(x_n)$?  Pick one and stick with it.  Personally, I despise the notation $\{x_n\}$ since a sequence is not a set, but that may be the standard notation in your course.
Also, the notation $\{x_n\}\in U$ isn't right: the sequence itself is not an element of $U$.  I would instead just say "a sequence $(x_n)$ in $U$" where the less precise English word "in" clearly conveys that you mean $x_n\in U$ for each $n$.
Finally, whichever notation you use, it is common to write just $x_n\to x$ with no brackets of any kind (though your course may have its own notation you're supposed to follow).


  
*Since $x\in V$ it implies there an neighborhood   $\rho_r(x) \in V$, which is completely contained in $V$. 
  

Here you have a similar notation error: it should be $\rho_r(x)\subseteq V$ instead of $\rho_r(x)\in V$.  You might also mention that you are using the assumption that $V$ is open in this step.


  
*This would mean that $U \cap V \neq \emptyset$. 
  

You could be more specific here, and say that $U\cap V$ is nonempty because $x_n\in U\cap V$ for any $n>N$.
Finally, while it is harmless to do so, it's not normal in mathematical writing to number the steps in a proof like you have done, and I don't think it makes your proof any clearer to do so in this case.  You can just write the steps one after another as sentences, organized into a paragraph or two.
A: Essentially we need to show that $\overline{U}\cap V=\emptyset$ and $U\cap\overline{V}=\emptyset$ where $U,V\subseteq\mathbb{R}$ are two open disjoint sets. Suppose otherwise i.e. $\overline{U}\cap V\neq\emptyset$ or $U\cap\overline{V}\neq\emptyset$. For simplicity consider only the case $\overline{U}\cap V\neq\emptyset$ (for the other situation the arguments runs exactly the same). This implies that there is some $x\in \overline{U}\cap V$. Therefore $x\in\overline{U}$ and $x\in V$. For the element $x$ to be in the closure of $U$ which is $\overline{U}$ it means that for all $\varepsilon>0$ the open ball $B_{\varepsilon}(x)\cap U\neq \emptyset$. On the other hand since $x\in V$ and $V$ is open by assumption then there exists some $\delta>0$ such that the open ball $B_{\delta}(x)\subset V$. In particular if we choose $\varepsilon=\delta$ then we get $B_{\delta}(x)\cap U\subseteq B_{\delta}(x)\subset V$ implying $U\cap V\neq\emptyset$. This contradicts that $U$ and $V$ are disjoint. 
A: This is true in any topological space $X.$  The definition of $\overline U$ is $\overline U=\cap F_U$ where $F_U$ is the family of all closed sets that each have $U$ as a subset.
Observe that $\overline U$ is a subset of any member of $F_U.$
If $V$ is open and disjoint from $U$ then $X\setminus V \in F_U$ so $\overline U\subset X\setminus V,$ which implies  $\overline U\cap V=\emptyset.$
Now interchange the letters $U,V.$
We can also say that the def'n of $\overline U$ implies that if $V$ is open then $V\cap \overline W=\emptyset$ for any $W\subset X$ such that $W$ is disjoint from  $V$ because  $X\setminus V\in F_W$ so $\overline W\subset X\setminus V.$ In particular when $W=U.$ And similarly with $U,V$ interchanged.    
