Why are these two expressions equivalent? Let $X$ be a topological space and let $A\subseteq X$. Can someone please tell me
1) what the difference between these two expressions is
2) if they are equivalent (if they aren't can someone give me two counterexamples in which in turn each expression holds and the other doesn't?)
3) how one would formulate the first expression in words ?

$ \forall x \in X: \ \ \forall (x_n)_n$ in $A:\ x_n \rightarrow x \ \Rightarrow x \in A$

$ \forall x \in X: \ \ \exists (x_n)_n$ in $A:\ x_n \rightarrow x \ \Rightarrow x \in A$

(The motivation behind this question is that in a course in topology which I took, the first expression was used in the following theorem that stated that $A$ was closed iff the first expresion would hold:

"Let $X$ be a first countable topological space, $A\subseteq X$ and $x \in X$.
Then we have: $A$ is closed $\Leftrightarrow$ $\forall (x_n)_n$ in $A,\ x_n \rightarrow x \ \Rightarrow x \in A$ "

And the second expression was used in the following theorem that stated that what the closure of $A$, $c(A)$, would look like:

"Let $X$ be a first countable topological space, $A\subseteq X$ and $x \in X$. Then we have: $c(A)= { x\in X |\  \exists (x_n)_n \textrm{ in }A:\ x_n \rightarrow x } $"

So my guess was, that the above expressions should also be equivalent formally (if we take the st theoretic notation from the second theorem away), since they express the same thing - so I think the proof of the equivalence should not use the fact that our topological space is actually first countable and one of the expression is identical to the fact that $A$ is closed. 

Please take note, that this is how the theorems were written down on the blackboard; that is all I have (even if it may be incorrect).

Since I'm not very good with formal notation I may have written the above expression falsely.)
The punctuation of the formulas in the question is very odd, and probably not enough to give a unique way of reading the formula.  But here is what the formulas mean if they are read with the usual conventions for quantifiers. 

The formula
$$
(\forall x \in X)(\forall (x_n) \in A)[ (x_n) \to x \Rightarrow x \in A]
$$
says that $A$ contains all its sequential adherent points. 

The formula
$$
(\forall x \in X)(\exists (x_n) \in A)[ (x_n) \to x \Rightarrow x \in A]
$$
says that for all $x \in X$ there exists a sequence $(x_n)$ of points of $A$ such that if $(x_n) \to x$ then $x \in A$. With the $T_1$ axiom this is simply equivalent to $A$ being nonempty:


If $A$ is nonempty, suppose I am given an $x \in X$. If $x \in A$ let $x_n = x$ for all $n$. Then $(x_n)\to x$ and $x \in A$, so the formula is true. Otherwise, $x \not \in A$. Fix some point $a \in A$ and let $x_n = a$ for all $n$. Then $(x_n) \to a$, so $(x_n) \not \to x$ by $T_1$, which means the formula is true. Either way, if $A$ is nonempty there is a sequence $(x_n)$ that works. 
If $A$ is empty, let $x$ be a point of the space. Then there is no sequence $(x_n)$ of points in $A$, so the formula is false. 




It is true that the first formula, 
$$
(\forall x \in X)(\forall (x_n) \in A)[ (x_n) \to x \Rightarrow x \in A]
$$
is equivalent to the following formula
$$
(\forall x \in X)[((\exists (x_n) \in A)[ (x_n) \to x]) \Rightarrow x \in A]
$$

This is an instance of a general scheme:
$$
((\exists z)\Psi) \Rightarrow \Phi) \Leftrightarrow (\forall x)(\Psi \to \Phi)
$$
which holds if $z$ is not mentioned in $\Phi$ (in this case $z$ is $(x_n)$). 

Perhaps the point I am making is that the parentheses are very important, because it makes a difference whether the quantifier is in the hypothesis of a conditional statement, or whether the entire conditional statement is in the scope of the quantifier.
If I'm not misunderstanding the correct meaning of your questions it seems to me quite obvious that they are not the same sequence in general but you have to clarify what you really mean with them.

In particular the second sentence, as I read it, means that every time you have a sequence of elements in $A$ converging at some point $x$ then this point is in $A$. This is true if you have a closed set but in general is not (take $1/n$ in $(0,1)$).

The first sentence (again as I read it) means that if every sequence in $A$ converges to the same point $x$ then this point is in $A$. This seems a little bit confusing because you can always find in a set which as more than one element two different sequence (the constant ones) converging to different points and so the hypotesis of your impications is almost always false.

---Edit---
From the new edit point of view is seems that the the sequences should be red as:


characterize closed set among others via convergent sequences: every time you've got a convergent sequence the limit is a point of the set in which the sequence lays.
characterize the closure of a set $A$. In particular you get the definition of closure (as you have said) taking the second $A$ in the sentence being $c(A)$.


Now, turning to the equivalence if you are asking about logic equivalence clearly there is a gap between the two sentences (not only referring to the $\forall$ and $\exists$ sa my previous answer was about): in particular you are comparing the second part of an equivalence (as in the characterization of a closed set) and a part of the definition of a set. So imo strictly logically speaking you cannot get an equivalence.

Topologically you have to care about separatedness axioms like said above and so on...

I hope I've undestood (better now) your question and that my answer is a little claryfing for you.