Closure of $A$ is closed. Definition: $C(A): A\cup A'$
Definition: $A'$: the set of limit points of $A$.
Exercise: Prove $C(A)$ is closed
My work: We need to show $C(A)^c$ is open. 
Let $x \in C(A)^c$, Then $x\notin A$ and $x \notin A'$, That implies $x$ is not a limit point of $A$.

Here my question: Why this happen?

Then exists a neighborhood $V$ of $x$ such that $V\cap A=\emptyset$
Hence, $C(A)^c$ is open. 

My other question: Why $V\cap A=\emptyset$ implies this:$C(A)^c$ is open
Thanks for all. I will be very grateful.
 A: first one: suppose not, then for any neighborhood around $x$, the intersection is nonempty. then $x$ is the limit point of $A$, a contradiction.
second one: once we get that neighborhood, pick an arbitrary point, say $y$ from $N(x)$, and $y\notin A$. then there exists a smaller neighborhood $N(y)\subset N(x)$. so $y \notin A'$. since y is arbitrary, we have proved that $N(x) \subset C(A)^C$, which is then open
A: What is your definition of openness? You can define it as: a set is open iff it is a neighbourhood of each of its elements.
Using that definition of openness:
If $x$ is not a limit point of elements in $A$, then there DOES NOT exist a sequence $(x_n)_{n\in\mathbb N}$ such that every neighbourhood of $x$ contains infinitely many points $x_n$.
This means (by logical negation), there EXISTS a neighbourhood $U$ of $x$ that contains at most finitely many elements of $A$. Call these elements that are both in $U$ and in $A$ from now on $x_{n_1},...x_{n_k}$ (for an appropriate $k\in\mathbb Z$).
Assuming you are in a T1 space (a space in which for every $x,y$ you can always find a neighbourhood of $x$ that does not contain $y$), you can separate $x$ from every single one of those finitely many points $x_{n_1},...x_{n_k}$ by neighbourhoods $V_1(x), ...,V_k(x)$. Since $k$ is finite, we can take the intersection of $V_1, ..., V_k$ and still end up with a neighbourhood of $x$. Therefore:
$$V=\bigcap_{i=1}^kV_i(x)$$
is a neighbourhood of $x$ that satisfies $V(x)\cap A=\emptyset.$
Since this holds for any $x\not\in C(A)$, we have shown that every $x\in(C(A))^c$ has a neighbourhood that is still completely within $(C(A))^c$ and therefore $(C(A))^c$ is neighbourhood of all its points (so it is open).
