Where does the notion of first countability is used? How does $f(\bar{A}) \subset \overline{f(A)}$ happened? Let $f:X\rightarrow Y.$If $f $ is continuous,then for every convergent sequence $x_n\rightarrow x$ in $X$,the sequence $f(x_n)$ converges to $f(x)$.The converse holds if $X$ is first countabe.
I was  getting  problem in proving the converse part but Convergence of $f(x_n)$ implies convergence of $x_n$ (From  Daniel Fischer's 2nd comment) cleared
my query upto very very extent.Now i wanted to know where does the  notion of first countability is used in proving its converse?
This is the proof given in Munkres
To prove the converse,assume that the convergent sequence condition is satisfied.Let A be as subset of $X$;We show  that $f(\bar{A}) \subset \overline{f(A)}$.If $x\in \bar{A},$there is a sequence converging to $x$.By assumption,the sequence $f(x_n)\rightarrow f(x) $.Since $f(x_n)\in f(A) \implies f(x)\in \overline{f(A)}$.Hence $f(\bar{A}) \subset \overline{f(A)}$.
How does $f(\bar{A}) \subset \overline{f(A)}$ happened?
My trial:

Let $x\in \bar{A} $ then $f(x)\in f(\bar{A})-----------------(1)$ .
Now we'll show that $f(x)\in \overline{f(A)} $. Since $x\in \bar{A}$
  so there exists a sequence,$<x_n>$ in $A$ such that $<x_n>\rightarrow
 x$ and by assumption we have $f(x_n)\rightarrow f(x)$ then  $f(x)\in
  \overline{f(A)}---------------------------(2)$
From (1) and (2) we have  $f(\bar{A}) \subset \overline{f(A)}$.

Is it correct?
 A: We want to show that $f(\bar A)\subset\overline{f(A)}$. That is, we want to show that if $x\in\bar A$ then $f(x)\in\overline{f(A)}$.
What's the definition of saying $x\in\bar A$? This is where we need first countability. Under the assumption of first countability, it's a fact that $x\in\bar A$ if and only if there is a sequence of points $x_n\in A$ such that $x_n\to x$.
Now, to show $f(x)\in\overline{f(A)}$, we need to show there is a sequence of points $y_n\in f(A)$ such that $y_n\to f(x)$. But using the assumption of the problem, since $x_n\to x$ we have $f(x_n)\to f(x)$, and we can then take $y_n=f(x_n)\in f(A)$, showing $f(x)\in\overline{f(A)}$, as desired.
A: Let it be that $f$ is not continuous at $x$. 
Then some open set $U$ exists with $f(x)\in U$ and such that $f^{-1}(U)$ is not a neighborhood of $x$. 
Now let $\mathcal B=\{B_1,B_2,\dots\}$ denote a countable neighborhood basis of $x$ with $B_1\supset B_2\supset\cdots$. 
$B_n\setminus f^{-1}(U)\neq\varnothing$ for every $n$ and picking some $x_n\in B_n\setminus f^{-1}(U)$ gives a sequence $(x_n)_n$ with $x_n\to x$ together with $f(x_n)\notin U$ for each $n$. Then we will not have $f(x_n)\to f(x)$.
The fact that $x$ has a countable neighborhood base makes it possible to construct such a sequence under the condition that $f$ is not continuous at $x$.
