$\overline{Y}=Y\cup \{v\in X| v ~ \text{is accumulation point of Y}\}$ 
Let $(X,\tau)$ be a topological space and $Y\subseteq X$. Show: $\overline{Y}=Y\cup \{v\in X| v ~ \text{is accumulation point of Y}\}$

It is $\overline{Y}=\bigcap\{A\subseteq X| A ~ \text{closed and}~ A\supseteq Y\}$
Our definition of an accumulation point of $Y$ is as follows:

We call $v\in X$ an accumulation point of $Y$, if in every neighborhood $N$ of $v$ there exists a point $y\in Y$ with $y\neq v$.

[Question: Does this mean that this $y$ is an element of that neighborhood as well? I misunderstood this at first and struggled to complete the proof...]
Proof:
Let $v\in Y\cup \{v\in X| v ~ \text{is accumulation point of Y}\}$.
We have to show that $v\in A$ for every closed $A$ with $A\supseteq Y$.
If $v\in Y$ there is nothing to show. So let $v\notin Y$. Then is $v$ an accumulation point of $Y$.
Suppose $v\notin A$. Then $v\in X-A$ open. In particular is $X-A$ a neighborhood of $v$. So there exists a $y\in X-A$ with $y\neq v$. Since $Y\subseteq A$ we have $y\in A$. Which is a contradiction to $y\in X-A$. 
So indeed $v\in A$ for every closed $A$ with $A\supseteq Y$.
And as desired $y\in\overline{Y}$.
Is this correct? 
I struggeld with the proof, because I misread the definition. I think at least. I did not noticed that the element of $Y$ has to be in that neighborhood, but I hope I am correct here.

Now let $v\in\overline{Y}$.

We have to show $v\in Y$ or $y\in \{v\in X| v ~ \text{is accumulation point of Y}\}$
Suppose the contrary $v\notin Y$ and $v\notin\{v\in X| v ~ \text{is accumulation point of Y}\}$
Let $A$ be closed with $A\supseteq Y$. Then $v\in A$, since $v\in\overline{Y}$.
$v$ is no accumulation point of $Y$. Hence it exists a neighborhood $N$ of $v$ and $y\in Y$ with $y=v$. 
But $v=y\notin Y$ by assumption. Contradiction.
So $v\in Y\cup\{v\in X| v ~ \text{is accumulation point of Y}\}$ and we have the equality of sets.
Is this proof correct?
Thanks in advance.
 A: A $v\in X$ is an accumulation point of $Y$ if for every neighborhood $N$  of $v$ there exists a point $y \in Y$ in $Υ$ so that $y \in N$ and $y\neq v$. So yes, $y\in N$.
Equivalently, a $v\in X$ is an accumulation point of $Y$ if for every neighborhood $N$  of $v$ we have 
  \begin{equation}\tag{1}
  Y\cap(N\setminus\{v\})\neq \emptyset
  \end{equation}
Set $Y'$ the set of all the accumulation points of $Y$.
Let's talk about your proof now.
  The first part of your proof is correct. In fact, I liked it! You just forgot to say "Let $A$ an arbitrary closed set with $A\supseteq Y$ " before "Suppose $v \notin A $", although you implied it in the end of the first part of your proof.
The second part of your proof is a little problematic. Notice that you never used crucially that $v \in \overline{Y}$. You also didn't use anywhere the closed set $A\supseteq Y$ and that $y \in A$ to reach contradiction. 
Furthermore, the part where you are saying "$v$ is not an accumulation point of $Y$, so there exists a neighborhood $N$ of 
  $v$ and $y\in Y$ with $y=v$" is not right.
For example, take $X=\mathbb{R}$, $Y=[1,2]$ and $v=0$. Then $0$ is not an accumulation point of $Y$: if it was, using the relation $Y\cup Y'\subseteq \overline{Y}$ that you proved in the first part, you should have  $0 \in Y'\subseteq Y\cup Y'\subseteq \overline{Y}=\overline{\big[1,2\big]}=[1,2]$, hence $0\in [1,2]$ which is a contradiction. Notice now that there exists no neighborhood $N$ of $v=0$ and $y\in Y=[1,2]$ with $y=v$, since $v=0<1\leq y$ for every $y \in Y$.
To prove that $\overline{Y} \subseteq Y\cup Y'$, or, equivalently, $\overline{Y}\setminus Y \subseteq Y'$,   you can use the fact that 
  \begin{equation}\tag{2}
  v\in \overline{Y} \iff N\cap Y \neq \emptyset \text{ for every neighborhood N of v}
  \end{equation}
Let $v \in \overline{Y}\setminus Y$ and a neighborhood $N$ of $v$. Since $v \in \overline{Y}$, using $(2)$ we see that $N\cap Y \neq \emptyset$. But $v \notin Y$, hence $Y=Y\setminus\{v\}$ and $Y\cap N=(Y\setminus\{v\})\cap N\neq \emptyset$, which means that  $Y\cap(N\setminus\{v\})\neq \emptyset$, and so using $(1)$ it follows that $v \in Y'$ .
