How does adding $0$ to the set $\mathbf A = \bigl\{\frac{1}{n}: n \in \mathbb N \bigr\}$ make it a closed set? By definition, a closed set is a set that contains its limit points. However, by the time the closed set contains its limit points, those points are no longer limit points and become isolated points. For example:
$\mathbf A = \{\frac{1}{n}: n \in \mathbb N \}$. The limit of this set (set $\mathbf A$) is clearly equal to $0$. This is because the $\epsilon$ -neighborhood $\mathbf V_{\epsilon}(0) \cap \mathbf A = \{\frac{1}{n} \}$, and $\frac{1}{n} \neq 0$. However, when $0$ is included, the $\epsilon$ -neighborhood $\mathbf V_{\epsilon}(0) \cap \mathbf A = \{0 \}$ for $\mathbf A=[0,\frac{1}{n} ]$. This will contradicts the definition of limit point of set A and hence $0$ must be an isolated point.
Another example: $\left(a,b\right)$ is an open interval with limit $a$ and $b$. Then its closure $\bar A$ will be $\left[a,b\right]$. By definition, the $\epsilon$ -neighborhood of any point in $\left[a,b\right]$ intersects the closure $\bar A$ at that same point, and hence, no points in that closure set is a limit point: A contradiction that closure sets are closed sets.
Also, I am trying to prove the lemma: If x is a limit point of $A \subseteq A'$, then x is limit point of $A'$.
Proof: Suppose x is a limit point of $A$, then there exists a sequence $(a_n)$ $\subset A \subseteq A'$: lim($a_n$)=x with $a_n$ $\neq x \forall n \in \mathbb N$. Then since $(a_n)$ $\subset A'$, it follows that x must be a limit point of $A'$.
So my questions are:
1. What is wrong with my contradiction in the 2 examples? Please explain them to me.
2. Is my proof for the lemma correct? I am going to use it for the proof that closure set is closed.
My background: I am studying elementary Real Analysis by starting with Abbot. I thank you very much for your help.
Extra question: We have this theorem: x is a limit point of set $A$ if and only if there exists a sequence $(a_n) \subset A$ such that $\lim (a_n)=x$ $\forall a_n \neq x$. So, for some finite $n \in \mathbb N$ such that $a_n = x$, x is still a limit point of set A . Is this correct? I thought that x would be an isolated points since we need $a_n \neq x \forall n \in \mathbb N$
I thank you again for your answers.
 A: Regarding the lemma: Suppose $x$ is a limit point of $A$. Then every neighborhood of $x$ contains a point in $A$ that is different than $x$. Since $A \subseteq A'$, then we can clearly see that these neighborhoods certainly contain a point in $A'$. The reason for this is because these neighborhoods contained at least a point in $A$, and we know $A$ is a subset of $A'$, so every point in $A$ is also in $A'$.
Using your proof: suppose there exists $(a_n) \rightarrow x$, and $(a_n)$ is a sequence in $A$. We know $a_n \neq x$ $\forall n \in \mathbb N$. Since $A$ is subset of $A'$, we must have that $(a_n)$ is a sequence in $A'$. Therefore, since we have the existence of a sequence in $A'$ that converges to $x$ where  $a_n \neq x$ $\forall n \in \mathbb N$, we must have that $x$ is a limit point of $A'$.
In your second example, remember a set is closed if and only if it contains all of its limit points. Your "contradiction" is wrong; $[a,b]$ contains its limit points (you mentioned that $(a,b)$ has limits $a$ and $b$), so therefore it is closed. You based your contradiction on the fact that "the neighborhood of any point in $\bar A = [a,b]$ will intersect $\bar A = [a,b]$ at that same point," but those same neighborhoods will also intersect other, distinct points in $[a,b]$. This means that every point in $[a,b]$ is a limit point, whereas you said no points were limit points. They all are.
Extra question: No, if the sequence has a term that is equal to $x$, then $x$ is not a limit point. The sequence of points must be in $A\backslash \{x\}$. The theorem is: "$x$ is a limit point of $A$ if and only if there exists a sequence in $A$ whose limit is $x$ and none of the terms in the sequence are equal to $x$". Or "$x$ is a limit point of $A$ if and only if there exists a sequence in $A\backslash \{x\}$ whose limit is $x$."
A: In the first case your neighbourhood of zero includes $\frac 1{n+1}, \frac 1{n+2} \dots$
In the second case I'm not sure what you mean. Any neighbourhood of a point in $[a,b]$ (we take $b\gt a$) intersects $[a,b]$ in (a set containing) an interval around the point. And any point within $[a,b]$ is a limit point.
A: First of all, recall that the notion of limit point has a meaning only for subsets  of the real numbers (or, more generally but not too generally, of a metric space).
A point $x$ is a limit point of $A$ if and only if, for every $\varepsilon>0$, the interval $(x-\varepsilon,x+\varepsilon)$ contains a point of $A$ different from $x$.
Equivalently, $x$ is a limit point of $A$ if and only if there exists a sequence $(a_n)$ in $A$, with $a_n\ne x$ for all $n$, and $\lim_{n\to\infty}a_n=x$.
This is linked to your extra question. If a finite number of terms of a sequence $(a_n)$ equal $x$, then there exists $k$ such that $a_n\ne x$ for all $n>k$. Then define $b_n=a_{n+k}$: the sequence $(b_n)$ has all terms different from $x$ and, if $(a_n)$ converges to $x$, also $(b_n)$ converges to $x$.
Let's look at your “contradictions”. Note that nowhere in the definition it is required that a limit point of $A$ doesn't belong to $A$.
For instance, every point of $[0,1]$ is a limit point of $[0,1]$. Indeed, if $0< x\le1$, let $n_0$ be the first natural number such that $1/n_0<x$. Then $x-1/n_0>0$ and the sequence $$a_n=x-\frac{1}{n+n_0}$$ is in $[0,1]$ and converges to $x$, without ever assuming the value $x$. For $0$, just consider $a_n=1/n$.
Similarly, $0$ is a limit point of $\{0\}\cup\{1/n:n\in\mathbb{N},n>0\}$. In this case, however, no other point is a limit point. So this set contains each of its limit points and is closed.
The important thing to note is that a sequence in $A$ converging to $x$ without ever assuming the value $x$ must exist, not that every sequence in $A$ that converges to $x$ shouldn't assume the value $x$. A sequence in $A$ showing that $x$ is a limit point is good for proving that $x$ is a limit point of the closure of $A$ as well!
Your proof of the lemma is good. In particular, it confirms that a limit point of $A$ is also a limit point of the closure of $A$.
