How to show that $f(S) \subset Y $ is dense, when $f$ is continuous and surjective, and $S \subset X$ is dense in $X$? Let $(X,d_x)$ and $(Y,d_y)$ be metric spaces. Furthermore, let $f:X \to Y$ be surjective and continuous. Furthermore: $S \subset X$ is dense in X. 
Question: How to prove that $f(S) \subset Y$ is dense in Y?
I wrote down the definitions of continuity:
$\forall x \in X, \forall a \in \mathbb{R} :  \exists \delta > 0 $ such that $\forall \epsilon > 0 : |x-a| < \delta \implies |(f(x) - f(a) | < \epsilon , $
and of $S \subset X$ being dense in X: 
$ \bar{S} = \{ x \in X | \forall \epsilon > 0 : \exists y \in S $ such that $d(x,y) < \epsilon \} = X $,
and of $f$ being surjective:
$\forall p \in Y : \exists x \in X : f(x) = p $. 
Using these definitions, I tried to prove: 
$\overline{f(S)} = \{ p \in Y | \forall \epsilon ' > 0 : \exists z \in f(S) $ such that $d(p,z) < \epsilon ' \} = Y.$ 
I couldn't figure it out, though. I tried proving this by contradiction, but to no avail. Could you please help me out? 
 A: Note that in a metric space the closure of a set is the set of limits of all sequences. In particular a set $A$ is dense in a space $X$ iff for any $x \in X$ there is a sequence $a_n \in A$ such that $a_n \rightarrow x$.
Hint: Let $y \in Y$. Then $y = f(x)$ for some $x \in X$. Now think about how to use the fact that $S$ is dense in $X$.
Edit: Since $S$ is dense in $X$ there exists a sequence $s_n \in S$ such that $s_n \rightarrow x$. Can you now find a sequence in $f(S)$ that tends to $y = f(x)$? (Remember, $f$ is continuous!)
Edit 2: To prove this using epsilon delta methods, take any $y \in Y$ and let $\epsilon > 0$. $f$ is surjective so $\exists x \in X$ such that $f(x) = y$. Since $f$ is continuous, $\exists\delta > 0$ such that $d_X(x,z) < \delta \implies d_Y(f(x),f(z)) < \epsilon$. Since $S$ is dense in $X$ there exists some $x_0 \in S$ such that $d_X(x,x_{0}) < \delta$. Hence $d_Y(f(x),f(x_{0})) < \epsilon$ and $f(x_0) \in f(S)$. As such $f(S)$ must be dense in $Y$.
A: Another characterisation of $S$ being dense in a metric space $X$ is the following:

for all $x \in X$ and all $\epsilon > 0$ there is a $y \in S$ with $d(x,y) < \epsilon$.

Given $v \in Y$ and $\epsilon > 0$, by surjectivity there is an $x \in X$ with $f(x) = v$, and by continuity there is a $\delta > 0$ such that $d_Y ( f(x) , f(y) ) < \epsilon$ for all $y \in X$ with $d_X (x,y) < \delta$. As $S$ is dense in $X$ there must be a $y \in S$ such that $d_X (x,y) < \delta$. But now $f(y) \in f [ S ]$ and by choice of $\delta$, $d_Y ( v , f(y) ) < \epsilon$.
A: Let $x=f(y)$ and $\epsilon>0$. We want to find $z\in f(S): d(y,z)<\epsilon$.
Since $f$ is continues exists $\delta>0: d(x,w)<\delta \Longrightarrow d(f(x),f(w))<\epsilon.$ 
Now use the fact that $S$ is dense in $X$ to find $w\in S:d(x,w)<\delta.$
A: Continuous functions preserve limits. Take a point in $x\in X$, and a sequence in $(x_n)\in S$ converging to $x$. Then $(y_n=f(x_n))\in f(S)$ will converge to $f(x)\in Y$ (by surjectivity). 
ADD Recall that in metric spaces, we can characterize density in the following manner:

A subset $D$ is dense on $X$ if for every $x\in X$ there exists a sequence of points $d_n\in D$ such that $d_n\to x$.

A: To show that $f(S)$ is dense in $Y$, we show that $f(S)\cap V\neq\emptyset$ for all non-empty open subsets $V$ of $Y$. By way of contradiction, suppose there is a non-empty open set $V\subseteq Y$ such that $f(S)\cap V=\emptyset.$ The preimage of $\emptyset$ under any function is $\emptyset$, so $$\begin{align}\emptyset &= f^{-1}\bigl(f(S)\cap V\bigr)\\ &= \bigl\{x\in X:f(x)\in f(S)\cap V\bigr\}\\ &= \bigl\{x\in X:f(x)\in f(S)\bigr\}\cap\bigl\{x\in X:f(x)\in V\bigr\}\\ &= f^{-1}\bigl(f(S)\bigr)\cap f^{-1}(V).\end{align}$$ We necessarily have $S\subseteq f^{-1}\bigl(f(S)\bigr)$, so it follows that $$S\cap f^{-1}(V)=\emptyset.$$ Now, $S$ is dense in $X$, and $f^{-1}(V)$ is open, as the preimage of the open set $V$ under the continuous map $f$, so we must have $f^{-1}(V)=\emptyset.$ But $V$ is a non-empty subset of $Y$ and $f$ maps surjectively to $Y$, so we have our contradiction.
A: Direct proof: let $U \subset Y$ be open and non-empty. Since $f$ is surjective, $f^{-1}(U) \subset X$ is non-empty, and open since $f$ is continuous so it preserves open-ness.
Now $S$ is dense in $X$ so $S \cap f^{-1}(U)$ is non-empty. So $\exists x \in f^{-1}(U) \cap S$ s.t. $f(x) \in U$, i.e. $\exists x \in S$ s.t. $f(x) \in U \implies f(S) \cap U$ is non-empty.
$\therefore f(S)$ is dense in $Y$.
