Prove that the Image of a Continuous Function on a Dense Set is Dense

I've been working on this for days and have made no progress, so any advice would be appreciated.

Definition: Let $\displaystyle R$ be an ordered field, and $\displaystyle Q \subset R$. $\displaystyle Q$ is dense in $\displaystyle R$ iff for any $\displaystyle x < y$ in $\displaystyle R$ there is a $\displaystyle q \in Q$ such that $\displaystyle x < q < y$.

Prove that, for any function $\displaystyle f$ continuous on a metric space $\displaystyle X$ and $\displaystyle E \subset X$ a dense subset, then $\displaystyle f(E)$ is dense.

So far what I have is hardly more than setting the problem up. Let $\displaystyle f(x), f(y) \in f(X)$. Thus $\displaystyle x \neq y$ (for now, lets assume $\displaystyle x < y$), and by density of $\displaystyle E$, $\displaystyle \exists e \in E$ such that $\displaystyle x < e < y $. I know that I will be basically done if I can show that $\displaystyle d(f(x), f(e)) < d(f(x), f(y))$ and $\displaystyle d(f(y), f(e)) < d(f(x), f(y))$. I also know that the inverse map must also be continuous. But that's all I got.

Thanks in advance.