Consider the following statement.
Let $M$ be a closed contact $3$-manifold with contact form $\lambda$ and resulting tangent plane distribution $\xi = \text{ker}(\lambda)$. Then there are no closed embedded surfaces $\Sigma \hookrightarrow M$ everywhere tangent to $\xi$.
I have a proof, but it seems a little too "high-tech". Namely, it is known that the $4$-manifold $\mathbb{R} \times M$ admits an exact symplectic structure $d(e^t\lambda)$, and furthermore a compatible almost complex structure $J$ that preserves the distribution $\xi$.
Then the embedding $\Sigma \hookrightarrow \{0\} \times M \subset \mathbb{R} \times M$ is a compact $J$-holomorphic curve. However, an exact symplectic manifold does not admit any embedded compact curves (one can compute that the Dirichlet energy of such an embedding with respect to the metric induced by the symplectic form and $J$ must be $0$).
Is there a quick way to see that this statement is true, only staying within the realm of contact topology? Furthermore, the above proof only works for $3$-dimensional contact manifolds. Do similar statements about Legendrian hypersurfaces hold for higher-dimensional contact manifolds?
$M$ being contact is the same as $\lambda \wedge d\lambda$ being a volume form.
One has $\lambda|_\Sigma = 0$. Since exterior derivative commutes with pullback/restriction, we get $d\lambda|_\Sigma = 0$ as well. It follows that $d\lambda$ vanishes on the distribution $\xi$.
If we take at a point a tangent frame $e_1$, $e_2$, $e_3$ where the latter two lie in $\xi$ and plug that into $\lambda \wedge d\lambda$, it evaluates to $0$ by our considerations. This is a contradiction since this is a volume form.
The same argument works in higher dimensions: A contact manifold of dimension $2n+1$ can not have a submanifold of dimension $>n$ everywhere tangent to the contact distribution.