Given a 2-form $\alpha$, I want to show that, if $X$ and $Y$ are vector fields such that $\alpha(X,Z)=\alpha(Y,Z)=0$ for all vector fields $Z$, then $\alpha([X,Y],Z)=0$ for all vector fields $Z$.
Since I had no better ideas, I just expanded everything out in a local chart, and I think it works. However, is there a nicer proof that sheds more light?
Edit: After seeing Jack Lee's answer below, I realized I forgot the crucial hypothesis that $\alpha$ is closed. I think that you can use $\alpha=d\beta$ locally, and $d\beta(X,Y)=X(\beta(Y))-Y(\beta(X))-\beta([X,Y])$ generally. Then, using the identities \begin{align*} d\beta(X,Z)=X(\beta(Z))-Z(\beta(X))-\beta([X,Z])&=0\\ d\beta(Y,Z)=Y(\beta(Z))-Z(\beta(Y))-\beta([Y,Z])&=0, \end{align*} we can expand out \begin{align*} d\beta([X,Y],Z)=[X,Y]\beta(Z)-Z(\beta([X,Y]))-\beta([[X,Y],Z]), \end{align*} replace $[X,Y]\beta(Z)$ with $X(Y(\beta(Z)))-Y(X(\beta(Z)))$, and apply the identities above to move the $Z$ to the outside. Also, we can use the Jacobi identity on the last term $\beta([[X,Y],Z])=\beta([X,[Y,Z]])+\beta([Y,[Z,X]])$ and again applying the identities above, I think everything ends up cancelling out. I don't know if there is supposed to be an easier way (at least this doesn't involve a bunch of indices).
This is not true. Take $\alpha$ to be the following $2$-form on $\mathbb R^4$ with coordinates $(w,x,y,z)$: $$ \alpha = dz\wedge dw - x \,dy \wedge dw, $$ and let $X,Y$ be the vector fields $$ X = \frac{\partial}{\partial x}, \qquad Y = \frac{\partial}{\partial y} + x \frac{\partial}{\partial z}. $$ Then $\iota_X\alpha = \iota_Y\alpha =0$, but $[X,Y] = \partial/\partial z$ and $\iota_{[X,Y]}\alpha = dw \ne 0$. (Here $\iota_X$ represents interior multiplication by $X$.)