how to prove the sequence based definition of a closure in metric spaces Let $A$ be a subset of a metric space $\Omega$. By definition, the closure of $A$ is the smallest closed set that contains $A$. How to prove that alternativelly, the closure is given by
(1): $\bar A = \{a_* \vert a_* = \lim_{n\rightarrow\infty} a_n;  \forall a_n \in A\}$ 
i.e. that $\bar A$ is given by limits of all converging sequences in $A$. 
I know how to prove that 
(2): A set $F\subset\Omega$ is closed iff the limits of all convergengent sequences from $F$ are in $F$. 
I feel that (1) and (2) should be related but how to prove (1)?  
All the help appreciated.
EDIT: I was thinking about trying this, but it is too engineering-like in spirit, I am not sure whether it would work. Take the set of all possible converging sequences which are obtained by taking elements in $A$, and call it $S$. 
$S = \{a_* \vert a_* \text{ as in Eq. (1)}\}$
Then there are two types of sequences, the ones that "saturate" i.e. where the same element starts repeating itself, call them $S_0$, and the ones that are "genuine" in the sense that all elements of the sequence are different, call it $S_*$. Thus
$S=S_0\cup S_*$
The saturating ones all represent $A$. Thus is should be that,
$S_0=A$
What is left, is the border $\partial A\equiv\bar A\setminus A$, and  this border should be somehow made of genuine sequences, i.e. one should somehow prove that
$S_* = \partial A$
e.g. I presume by exploiting (2). Would a proof based on this strategy be possible? 

Consider the set $X=\{x\,|\,a_n\to x;\;\forall n\, a_n\in A\}$.
Clearly $A\subset X$. You should be able to prove that $X\subset\Omega$ is closed. Now suppose that $Y\subset\Omega$ is closed, $Y\ne X$ and $A\subset Y\subset X$. Then there must be some $x\in X$ such that $x\ne Y$. Work from there to show that no such $Y$ exists and then you are done.
