I am trying to prove this
For any vector bundle $\pi : E \rightarrow M$ and any $x \in M$:
$\forall v \in E_x \exists s \in \Gamma(E)$ s.t $s(x) = v$
$\Gamma(E)$ is the space of (smooth) sections of the vector bundle
Unfortunately I have nothing really useful to put here, I haven't gotten anywhere, plus I don't have much experience with diff geometry proofs. I would be nice if someone could show me how it is done as an example.
Let $x \in M$ and $v \in E_x$ be arbitrary. First, let $E \cong M\times \mathbb{R}^n$ be trivial. Then $v \cong (x, v')$. Take the "constant" section $s: M \to E, \, p \mapsto (p,v')$.
Now let $E$ be arbitrary. Then, by the above, there is some neighborhood $U\subseteq M$ of $x$ and a local section $s': U \to E$ with $s'(x) = v$. Take a smooth bump function $f: M \to \mathbb{R}$ with $\text{supp}f \subseteq U$ and $f(x) = 1$. Define $$s: M \to E, \, s(p) = \begin{cases} f(p)s'(p), & p \in U \\ 0_p, & p \in M \smallsetminus \text{supp} f \end{cases}$$ where $0_p \in E_p$ is the zero-vector. Then $s$ is well-defined, smooth on the open sets $U$ and $M \smallsetminus \text{supp} f$ and $M = U \cup (M \smallsetminus \text{supp} f)$, i.e. $s \in \Gamma(E)$. Furthermore, $s(x) = f(x)s'(x) = v$.