Obviously, this is a question that could be interpreted in different ways. For me, Darboux's theorem is the symplectic analogue of the theorem that a flat Riemannian manifold (i.e. one where Riemann's curvature tensor vanishes) is locally the same as $\mathbb{R}^n$. For Darboux, the analogue of the Riemann curvature tensor is $d\omega$, the differential of the symplectic form.

So one variation on your question is "why should the symplectic form on phase space be closed?" which has a very clear answer:

If $d\omega\neq 0$, then the form $\omega$ is not invariant under time translation! That is not really very physically realistic.

The way to compute this is by taking the Lie derivative of $\omega$ by a Hamiltonian vector field $X_f$, using Cartan's magic formula $\mathcal{L}=d\iota+\iota d$. One finds that $$\mathcal{L}_{X_f}\omega=d(\omega(X_f,-))+d\omega(X_f,-,-)=ddf+d\omega(X_f,-,-))=d\omega(X_f,-,-),$$ so closedness is essentially equivalent to invariance of $\omega$ under time translation.

Obviously, this is a question that could be interpreted in different ways. For me, Darboux's theorem is the symplectic analogue of the theorem that a flat Riemannian manifold (i.e. one where Riemann's curvature tensor vanishes) is locally the same as $\mathbb{R}^n$. For Darboux, the analogue of the Riemann curvature tensor is $d\omega$, the differential of the symplectic form.

So one variation on your question is "why should the symplectic form on phase space be closed?" which has a very clear answer:

If $d\omega\neq 0$, then the form $\omega$ is not invariant under time translation!

That is definitely not very physically realistic.

The way to compute this is by taking the Lie derivative of $\omega$ by a Hamiltonian vector field $X_f$, using Cartan's magic formula $\mathcal{L}=d\iota+\iota d$. One finds that $$\mathcal{L}_{X_f}\omega=d(\omega(X_f,-))+d\omega(X_f,-,-)=ddf+d\omega(X_f,-,-))=d\omega(X_f,-,-),$$ so closedness is essentially equivalent to invariance of $\omega$ under time translation.