Darboux's theorem

Step 1 (set up the linear problem). By a linear change of coordinates one first arranges that at the chosen point $p_0$, the form $\omega$ equals the standard form $\omega_0$ exactly at $p_0$ itself, since every non-degenerate skew-symmetric bilinear form on $\mathbb{R}^{2n}$ can be brought to standard form by a suitable basis.

Step 2 (interpolate). Define the family $\omega_t=(1-t)\omega_0+t\omega$ for $t\in[0,1]$. Since $\omega_0$ and $\omega$ agree at $p_0$ and both are closed and non-degenerate there, $\omega_t$ is also closed and, after shrinking the neighbourhood if necessary, non-degenerate for every $t$.

Step 3 (solve Moser's equation). Because $\omega$ minus $\omega_0$ is closed and vanishes at $p_0$, the Poincaré lemma produces a 1-form $\sigma$ with $d\sigma=\omega-\omega_0$. Moser's trick looks for a time-dependent vector field $X_t$ solving $\iota_{X_t}\omega_t=-\sigma$; since $\omega_t$ is non-degenerate this determines $X_t$ uniquely at every point.

Step 4 (integrate the flow). Let $\varphi_t$ be the flow generated by $X_t$. A direct computation with Cartan's formula gives $\mathcal{L}_{X_t}\omega_t=d\iota_{X_t}\omega_t+\iota_{X_t}d\omega_t=-d\sigma=-(\omega-\omega_0)$, combined with $\dot\omega_t=\omega-\omega_0$, so that $\frac{d}{dt}(\varphi_t^*\omega_t)=0$: the pullback $\varphi_t^*\omega_t$ is constant in $t$ and equals $\omega_0$ for every $t$.

Step 5 (conclude). Setting $t=1$ gives $\varphi_1^*\omega=\omega_0$ on the shrunk neighbourhood, so the coordinates pulled back along $\varphi_1$ are the desired Darboux coordinates.