I want to show that there exists a symplectic vector field on the $2n$ torus $\mathbb{T}^{2n}$, endowed with the unique symplectic form $\omega$ that pullsback to the canonical symplectic form $\omega_0$ on $\mathbb{R}^{2n}$ under the quotient map $\pi:\mathbb{R}^{2n}\to\mathbb{T}^{2n}$, which is not Hamiltonian. We identify $T_x\mathbb{T}^{2n}\cong\mathbb{R}^{2n}$ for all $x\in\mathbb{R}^{2n}$ and we define the vector field $X\in\mathcal{X}(\mathbb{T}^{2n})$ by $X(x)=v$ for some fixed $0\neq v\in\mathbb{R}^{2n}$. Then I want to show that $d\iota_X\omega=0$.
I consider the vector field $\tilde{X}$ on $\mathbb{R}^{2n}$ defined by $\tilde{X}(x)=v$. Then $\tilde{X}$ is symplectic and satisfies $d\pi_x\tilde{X}(x)=X_{\pi(x)}$, so $d\iota_{\tilde{X}}\omega_0=0$. But $$ d\iota_{\tilde{X}}\omega_0=d\omega_0(\tilde{X},\cdot)=d(\pi^*\omega(\tilde{X},\cdot))=d\omega_{\pi(\cdot)}(d\pi\tilde X,d\pi\cdot)\underbrace{=}_{?}d\omega(X,\cdot)=d\iota_X\omega, $$ so $X$ is symplectic. I am not sure about the step with a question mark, as the $d\pi$ in the second argument disappears. Is there any justification for this?
Now, I want to show that $X$ is not Hamiltonian. As always, we assume it is, so there exists a smooth map $H:\mathbb{T}^{2n}\to\mathbb{R}$ such that $\iota_X\omega=dH$. But I don't see how to arrive at some contradiction now.
Hint: On a compact manifold, an Hamiltonian vector field has a fixed point. Let $H$ be the Hamiltonian, there exists $x$ such $H(x)$ is a maximum, and $dH(x)=0=i_{X(x)}\omega_x$ implies that $X(x)=0$.
On $\mathbb{T}^n$ take any vector field induced by $\phi_t(x)=x+ta, a\in\mathbb{R}^{2n}-0$, $\phi_t(x)$ is a symplectic vector of $\mathbb{R}^{2n}$ which induces an symplectic vector on $\mathbb{T}^{2n}$.