This may be a slightly vague/ill-posed question. I apologize in advance.
Suppose we have a symplectic manifold $(M,\omega)$ and a smooth submanifold $U$.
Suppose at any point $m \in U$ we have that the orthogonal complement (with respect to the symplectic form) to $T_m(U)$ in $T_m(M)$ is zero. Is $U$ necessarily open?
Is this because we can show $\dim U = \dim M$ or something?
Yes, This follows from linear algebra. For a vector space $V$ with a $2$-form $\omega$ (i.e. a skew-symmetric, non-degenerate bilinear map), for each subspace $W \subset V$ let $W^{\omega}$ denoted the orthogonal complement with respect to $\omega$ (i.e. $W^{\omega} = \{y \in V| \space \omega (x,y) = 0 \space\space \forall x \in W\}$).
Then in general it will not be true that $V = W \oplus W^{\omega}$(as is the case for Riemannian metrics for example). However, we do have the formula:$$\dim(W) + \dim(W^{\omega}) = \dim(V). $$
This is proven in Lemma 2.2 of Introduction to symplectic topology by McDuff and Salamon. Now we return to the problem at hand. Set $V = T_{m}(M)$ and $W = T_{m}(U)$. Then by assumption $W^{\omega} = \{0\}$, hence $W$ has the same dimension as $V$, hence we must have $W=V$. This implies that $\dim(U) = \dim(M)$. Now it follows that $U$ is open.