I want to show that
$$(\forall f\in X^*: f|_F \equiv 0 \Rightarrow f\equiv 0) \Leftrightarrow F \text{ dense in }X.$$
I have proven the direction $\Leftarrow$. But I have problems with the other direction. If this direction does hold, we could equivalently show that if $F$ is not dense in $X$ the other statement doesn't hold.
If I just think of finding an $f:X\to\mathbb{R}$ that has to be continuous and $f|_F \equiv 0$ this is simple. Since $\bar{F} \neq X$ and $\bar{F}$ closed it holds that $\emptyset \neq X \setminus \bar{F}$ open and so we can construct a continuous functional that is not $0$ in an open ball around an $x\in X \setminus \bar{F}$.
However such a functional $f$ obviously has not to be linear. And for example in the case of $X = \mathbb{R}$ and $F=\mathbb{Z}$ there doesn't even exist such a functional.
So I don't really have an idea to prove this direction. I'm thankful for any hints on this.