Showing that vector field $X$ defined by $X_p := \widehat{\omega_p}^{-1}(\alpha_p)$ is smooth This is exercise 1, p.22 in the book Lectures on Symplectic Geometry by Ana Cannas da Silva.

Let $M$ be a smooth manifold and $\alpha \in \Omega^{1}(M)$ such that
  $\omega = -d\alpha$. Then there exists a unique vector field $X \in
\mathfrak{X}(M)$ such that the interior multiplication of $X$ with
  $\omega$ is $-\alpha$, i.e. $i_X\omega = -\alpha$.

Uniqueness is clear from the fact that $\widehat{\omega_p} : T_pM \to T^*_pM$ defined by $\widehat{\omega_p}(v) := \omega_p(v,\cdot)$ is an isomorphism due to the symplectic structure.
Existence. Let $p \in M$. Then we have to solve $i_X\omega = -\alpha$ which is equivalent to $\widehat{\omega_p}(X_p) = -\alpha_p$ or $$X_p = -\widehat{\omega_p}^{-1}(\alpha_p)$$ So left to show is that $X \in \mathfrak{X}(M)$. Does anyone see why the mapping $p \mapsto X_p$ is smooth?
Edit. One can check that $\widehat{\omega}: TM \to T^*M$ is a smooth bundle isomorphism. Hence we define $$X := -\widehat{\omega}^{-1}(\alpha).$$ Let us check that $X \in \mathfrak{X}(M)$. Clearly, $X$ is a section by definition of $\widehat{\omega}$ and furthermore smooth since it is the composition of smooth mappings.

There might be a more elegant way to prove this. But you could just apply Darboux's theorem and get coordinates $(p^i, q^i )$ around any point such that
$$\omega = \sum_i \mathrm d q^i \wedge \mathrm d p^i$$
in that coordinate neighborhood. In that case, $$X = \sum_i (X^{p^i} \partial_{p^i} + X^{q^i} \partial_{q^i})$$ (analogous for a vector field Y which I'm going to use in a blink of an eye) and
$$i_{X} \omega (Y) = \sum_i X^{p^i} i_{\partial_{p^i}} (\sum_j \mathrm dq^j \wedge \mathrm d p^j )(Y) + \sum_i X^{q^i} i_{\partial_{q^i}} (\sum_j \mathrm dq^j \wedge \mathrm d p^j )(Y) = \dots = \sum_i (- X^{p^i} Y^{q^i} + X^{q^i} Y^{p^i}) \stackrel{!}{=} \sum_i (\alpha_{p^i} Y^{p^i} + \alpha_{q^i} Y^{q^i})$$
But from that you can infer that $$X= \sum_i (\alpha_{p^i} \partial_{q^i} - \alpha_{q^i} \partial_{p^i})$$
The components are obviously smooth functions of the coordinates, hence this vector field is smooth.
This is independent of the condition that $\mathrm d \alpha = - \omega$. It's a general fact that a symplectic form turns smooth 1-forms into smooth vector fields and vice versa.
