Let $(M,\omega)$ be a symplectic manifold.
Being $\omega$ a non-degenerate $2-\textrm{form}$ on $M$, for any $f\in C^{\infty}(M)$ there exists a unique $X_f\in\textrm{Lie}(M)$ such that $df=i(X_f)\omega$. The map $f\in C^{\infty}(M)\to X_f\in\textrm{Lie}(M)$ is obviously $\mathbb{R}-\textrm{linear}$.
We introduce the Poisson bracket over $C^{\infty}(M)$ by defining $\{f,g\}=X_f(g)\equiv \omega(X_g,X_f)$, for any $f$ and $g$ smooth function on $M$.
Immediatly by definition $\{\cdot,\cdot\}:C^{\infty}(M)\times C^{\infty}(M)\to C^{\infty}(M)$ is an antisymmetric $\mathbb{R}-\textrm{bilinear}$ map.
The Jacobi identity for the Poisson bracket can be easily rewritten as (*)$X_{\{f,g\}}=[X_f,X_g]$, for any $f$ and $g$ smooth functions on $M$. So $C^{\infty}(M),\{\cdot,\cdot\})$ is a Lie algebra if and only if the map $C^{\infty}(M),\{\cdot,\cdot\})\ni f\to X_f\in\textrm{Lie}(M)$ is a homomorphism of $\mathbb{R}-\textrm{algebras}$.
Being $d\omega=0$, by the H.Cartan's formula we get that a smooth vector field $X$ on $(M,\omega)$ is symplectic, i.e. $\mathcal{L}(X)(\omega)=0$, if and only if it is locally hamiltonian, i.e. $d.i(X)\omega=0$.
Now the condition (*) is a consequence of the much more strong statement: if $Y$ and $Z$ are symplectic vector fields on $(M,\omega)$, i.e. $\mathcal{L}(Y)(\omega)=\mathcal{L}(Z)(\omega)=0$, then $[Y,Z]=-X_{\omega(Y,Z)}$, i.e. $[Y,Z]$ is a hamiltonian vector field with $-\omega(Y,Z)$ as Hamilton function. Proof. $i([Y,Z])\omega=\mathcal{L}(Y).i(Z)\omega-i(Z).\mathcal{L}(Y)\omega=d.i(Y).i(Z)\omega+i(Y).d.i(Z)\omega=d(\omega(Z,Y))$. (Having used the hypothesis, the H.Cartan's formula, and the formula $[\mathcal{L}(Y),i(Z)]=i([Y,Z])$).
Probably, at this time and after Jose's answer, my answer could be not useful for you. But anyway I tried to find a coordinate-free proof of the Jacobi identity for the Poisson bracket over a symplectic manifold, in order to test my own comprehension. And now I try to write it here as an exercise in exposition and hoping to be useful to casual readers.