Continuous function between metric spaces Let $(X , d)$ and $(Y , \rho)$ be two metric spaces. Let $S = \{x_1, x_2, x_3, ...\}$ be a countable dense subset of $X$. Let $f : X \rightarrow Y$ be a continuous function. The prove that; 
For a closed set $F$ in $Y$,  $f(x)$ belongs to the closed set $F$ if and only if for each $n \in \mathbb{N}$, there is some $s \in S$ with $d(x, s) < \frac{1}{n}$ and $ρ(f (s), F) < \frac{1}{n}$. 
Thanks in advance.

"$\Rightarrow$" Let $f(x) \in F$, i.e. $x \in f^{-1}(F)$. Let $n \in \mathbb{N}^+$. Since $f$ is continuous at $x$, we find some $\delta>0$ such that $d(x,y) < \delta $ implies $d(f(x),f(y))< \frac{1}{n}$. Since $S$ is dense, there is some $s \in S$ such that $d(x,s) < \min(\delta,\frac{1}{n})$. Hence, $d(x,s) < \frac{1}{n}$ and $d(A,f(s)) \leq d(f(x),f(s)) < \frac{1}{n}$.
"$\Leftarrow$" It suffices to prove that $f(x) \in \overline{F}$, i.e. that for every $n \in \mathbb{N}^+$ there is some $y \in F$ such that $d(f(x),y)<\frac{1}{n}$. Since $f$ is continuous, there is some $m \in \mathbb{N}^+$ such that $d(x,y) < \frac{1}{m}$ implies $d(f(x),f(y)) < \frac{1}{2n}$. We may assume $m \geq 2n$. By assumption there is some $s \in S$ such that $d(x,s) < \frac{1}{m}$ and $d(f(s),F) < \frac{1}{m}$. Choose $y \in F$ such that $d(f(s),y) < \frac{1}{m}$. Then
$d(f(x),y) \leq d(f(x),f(s)) + d(f(s),y) < \frac{1}{2n} + \frac{1}{2n} = \frac{1}{n}.$
