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In da Silva's Lectures in Symplectic Geometry page 105, it was claimed that if $\omega$ is nondegenerate symplectic 2-form then there exists unique vector field $X_H$ such that $\iota_{X_H}\omega=dH$ for some smooth function $H$.

I cannot quite see this intuitively. Naively, this seems to be a fair claim since at least $d^2H=0$ showing that indeed $\omega$ is closed. But I cannot see the principle behind it: is this a statement of Poincar\'e lemma? That this is true is somewhat remarkable to me, unless I am misunderstanding differential forms in general to miss this obvious statement (I think it is not obvious).

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    You seem to be getting a few things wrong: $d^2H=d\iota_X \omega=0$ does not imply that $d\omega=0$, and this does not have anything to do with the Poincaré lemma. – 2017-02-09
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    @Danu yes I think I have phrased this badly due to some misunderstanding. I will edit this question at some point if I think I am still stuck. – 2017-02-09

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Non-degeneracy means the map

$$\omega^{\sharp} : TM \to T^* M$$ $$\omega^{\sharp}(X)(Y) = \omega(X, Y)$$

is an isomorphism (hence it's also an isomorphism on sections). So you just need to set $X_H = (\omega^{\sharp})^{-1}(dH)$.

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    is this somewhere in the lectures? I am actually jumping directly to Chapter 18 from Chapter 2 since I am reading this with physics as main objective, so if it is somewhere in between could you refer me to it? Also, your solution seems to imply that I can just directly choose $dH$ instead of other one-forms. Why? – 2017-02-09
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    This is a decent source: https://www.dpmms.cam.ac.uk/study/IB/LinearAlgebra/2008-2009/bilinear-08.pdf for the background assumed in Pedro's answer. – 2017-02-09