Linear functional f(v)=0 imply v=0

Good day, I want to prove the following theorem:

1) For any nonzero vector $ v \in V $, there exists a linear funtional $ f \in V^*$ for wich $f(v) \neq 0 $

I know that if $f$ is a lineal functional then we have 2 posibilities

1)$\dim \ker(f)=\dim V$

2)$\dim \ker(f)=\dim V-1$

I've tried to suppose that, for all $v \neq 0 $ and $f \in V^*$ We have $f(v)=0$,

Then $Im(f)=0$ for all $f \in V^*$ and $ v \in V$ but then the only linear functional is $f=0$ ($V^*=\{0:V \to F\}$) and from this follows: $\dim \ker(f)=\dim V$. But 2) suggest the existence of a linear functional which $\dim \ker(f)=\dim V-1$, so ¿is that the contradiction?

Let $v\in V, v\ne0$ and let $H$ a subspace of $V$ such that

$$V=\operatorname{span}(v)\oplus H$$ Now define a linear functional $f$ on $V$ by

$$f(v)=1\quad\text{and}\quad f(h)=0\;\forall h\in H$$ so we proved that for all $v\ne0$ there is $f\in V^*$ such that $f(v)\ne0$ and then by contrapositive we have the desired result.