is continuous image of dense set dense? Here is the problem:

Let $f : X \to Y$ be a continuous map between topological spaces.  Let $E$ be a dense subset of $X$ (that is, $\operatorname{Cl} E = X$), where $\operatorname{Cl}E$ represents the closure of $E$.  Prove: $\operatorname{Cl} f(E) = f(X)$

I know continuity of $f$ implies $f(\operatorname{Cl} E) \subseteq \operatorname{Cl} f (E)$, but I have no idea to show the converse.
$f[X] = f[\operatorname{cl}(E)] \subset \operatorname{cl}(f[E])$. This implies that every point of $f[X]$ is in the closure of $f[E]$, i.e. $f[E]$ is dense in $f[X]$.

Alternatively from the above inclusion $f[X] \subseteq \operatorname{cl}(f[E])$ we see: $$f[X]\supseteq \operatorname{cl}_{f[X]}(f[E]) = \operatorname{cl}(f[E]) \cap f[X] \supseteq f[X] \cap f[X]=f[X]$$ and hence $\operatorname{cl}_{f[X]}(f[E]) = f[X]$ which says $f[E]$ is dense in $f[X]$.
No. Take constant function on R to R.