If $\theta \in \frak{X}^* \mathrm{(M)}$ and $\theta (X) = 0$ $\forall X \in \frak{X} \mathrm{(M)}$ then $\theta = 0$.
How do I prove this statement?
Consider a manifold $M$ with chart $x^1, \dots, x^n$.
Assume $\theta (X) = 0$ $\forall X \in \frak{X} \mathrm{(M)}$.
Since $\theta (X) = \sum \limits_{i=1}^{n}{\theta_i X^i} $ then $\theta (X)|_p = \sum \limits_{i=1}^{n}{\theta_i|_p X^i|_p}$ for any $p\in M$.
Is it now okay to say that since we assumed $\theta (X) = 0$ $\forall X \in \frak{X} \mathrm{(M)}$, then $\theta (X)|_p = \sum \limits_{i=1}^{n}{\theta_i|_p X^i|_p} = 0$ $\forall X^i|p \in \mathbb{R}$, so for the equation to hold true, $\theta_i|p = 0$ $\forall p \in M$ which means $\theta = 0$?
Yes, what you say is correct. Recall that at each point of $M$ a one form is just an element of $T_pM^*$. You have proved that its value when acting on any vector of $T_pM$ is zero, so in particular this is true for a basis of $T_pM$. But a linear form is totally determined by its value on a basis, so $\theta_p$ must be the zero 1-form. Since this holds for all $p$ $\theta $ is identically zero.