$f(x_0)=0 \forall f \in X^*$ then $x_0=0$. Let X be a vector space of infinite dimension (possibly) and $x_0 \in X$.
I would like to show that if $f(x_0)=0 \forall f \in X^*$ then $x_0=0$.
So the first intuition would be to pick a function such as $Id_X$ but this is not even into the dual $X^*$.
Edit : 
Actually $X^*$ was the topological dual, so containing the continuous functions (Sorry, I totally forgot that this could be confused with the algebraic dual, but actually, the answers still helped me for this, so thanks).
 A: I think that the statment is not necessarily true without assuming the axiom of choice. But if we assume it, then $X$ has a basis. It's easy to dedudce from this that, if $x_0\neq0$, then $X$ has a basis $B$ such that $x_0\in B$. Now, consider $\varphi\in X^*$ which maps $x_0$ into $1$ and all other elements of $B$ into $0$.
A: I take the hypothesis that $X^*$ is the algebraic dual space as you don’t mention any norm or something similar.
Suppose that $x_0 \neq 0$ and complement it with $(y_i)_{i \in I}$ to get a basis $\mathcal B=(x_0,(y_i)_{i \in I})$. This uses axiom of choice.
Then define $f \in X^*$ on that basis by $f(x_0)=1$ and $f(y_i)=0$ for $i \in I$.
You’re done.
A: The question does not specify any topology, so it could well be that $X^*$ is supposed to be the algebraic dual of $X$. As a matter of interest, if $X$ is a Banach space and $X^*$ is supposed to be the space of continuous linear functionals, then the construction in the other two answers, "choose a basis and send one basis element to $1$, the others to $0$", does not necessarily give a continuous functional:
Say $\Lambda$ is an unbounded linear functional on $X$ and let $Y$ be the nullspace of $\Lambda$. Let $B_0$ be a Hamel basis for $Y$, and let $B=B_0\cup\{x_0\}$ where $\Lambda x_0=1$. Then $\Lambda$ is the functional constructed above.
So now we know: Coordinate functionals with respect to a Hamel basis need not be continuous.
