Let $(M, \omega)$ be a symplectic manifold, $M' \subset M$ be a submanifold of $M$. Suppose that for every $x \in M'$, there exist a symplectic subspace $V_x$ of $T_xM$ such that $T_xM = T_xM' \oplus V_x$ and $T_xM'$ is symplectically perpendicular to $V_x$.
Can we deduce from this that $M'$ is also symplectic ?
My motivation to ask this question is to understand the proof of the cross-section theorem (theorem 3.3 in the paper Non abelian convexity by symplectic cuts) which seems to use an argument similar to what I'm asking for!
Recall from symplectic linear algebra: if $W\subset(V,\omega)$ is a symplectic subspace, then also its symplectic orthogonal $W^{\omega}$ is a symplectic subspace. Indeed, $$ (W^{\omega})^{\omega}\cap W^{\omega}=W\cap W^{\omega}=\{0\}, $$ using in the last equality that $W\subset(V,\omega)$ is symplectic.
By assumption you have $T_{x}M'=(V_{x})^{\omega}$, where $V_{x}$ is a symplectic subspace of $T_{x}M$. Hence also $T_{x}M'$ is a symplectic subspace, and therefore $M'$ is a symplectic submanifold of $(M,\omega)$.