Is every point of every open set $E \subset R^2 $ a limit point of E?(Proof-verification) NOTE: There are some other similar questions, but I got a negative answer to this question from my proof. Please find out the errors in my reasoning. 
$\mathbf {Claim:}$ Is every point of every open set $E \subset R^2$ a limit point of E? Answer the same question for closed sets in $E \subset R^2$
From "Baby Rudin"
$\mathbf {Proof:}$ $\emptyset$ is both open and closed in every topological space. $R^2$ is a metric space, which is a kind of topological space, so $\emptyset$ is both open and closed in it. $\emptyset$ has no limit point because its neighborhood has no other point to include, so we can get a negative answer to both questions.
If possible, please have a look at the two questions that I got while reading the answer to this similiar question:
Proof that every point of every open set E⊂ℝ^2 is a limit point of E?⊂ℝ2-is-a-limit-point-of-e


*

*"$q_s=(x_1+s,x_2)$" should be $q_s=(x_1+s,y_1)$, right?

*I still don't understand why there should be $\epsilon$. Why can't r complete the proof?


I don't have the right to comment on the original post led by the above link, so I ask the two questions here.
Finally, I find this forum very active, responsive and helpful, but not quite friendly to newcomers.
 A: Every k is A $\iff $ there is no k that is not A.
So every point of the empty set is a pink alligator that eats square circles... because the empty set does not have any points that are not pink alligators that eat square circles...  because the empty has no points at all so none of them can avoid being a pink alligator that eat square circles.
To the statement is true for the empty set.  Every point in $\emptyset$ has property $X$ because there are no points in $\emptyset$ that don't have property $X$.  
Every point is a limit point.  And every point is not a limit point.  And every point is green.  And every point is not green.  This is because there are zero points all "all of zero" is .... zero.  And zero of the points are limit points. So all points are limit points because zero is all the points there are and zero are limit points.
So that is not a counter example.
1)  Yes, that was a typo.
2)Because $r$ defines one neighborhood.  We must prove this is true for all neighborhoods.
A: For the open set part, you are asked to show whether the following statement is correct:
Let $E$ be an open set in $\mathbb{R}^2$. If $p\in E$, then $p$ is a limit point of $E$.
Your counterexample by letting $E=\emptyset$ is not correct. Because by definition, empty set contains no element, which means you can't select element from it, hence the if part of the above statement is false. Then for the statement $P\rightarrow Q$, if $P$ is false, no matter whether $Q$ is true or false, we say $P\rightarrow Q$ is true. Therefore, for $E=\emptyset$, we know the statement is true.
For the proof in your link:


*

*Yes. $q_s=(x_1+s,y_1)$.

*We are asked to show $p$ is a limit point of $E$, which means for any $\epsilon>0$, we need to find a $q_\epsilon\in E\cap B(p,\epsilon)\setminus\{p\}$.
A: You can do this more easily by noting that $\mathbb R^2$ has no isolated points. A  point $p$ in a space $X$ is isolated iff $\{p\}$ is open. 
Let $p\in U\subset R^2$ where $U$ is open.   If $V$ is any nbhd of $p,$ then there exists open $V'$ with $p\in V'\subset V.$ 
Now  $U\cap V'$ is open and not empty (as it contains $p$) so $\phi \ne U\cap V'\ne \{p\}.$ So there exists $q\in U\cap V'$ with $q\ne p.$
So any nbhd $V$ of  $p$ contains a point $q$ with $p\ne q\in U.$ 
