A smooth family of symplectic forms Let $A(t)\in\mathbb R^{2n\times 2n}$ be a smooth family of nondegenerate skew-symmetric matrices, $t\in\mathbb R$. Then the family $A(t)$ represents the family of symplectic forms $\omega_t(u,v)=\langle A_tu,v\rangle$. Is it possible to construct a smooth family $S(t)=\{u_1(t),\ldots,u_n(t),v_1(t),\ldots,v_n(t)\}$ of bases in $\mathbb R^{2n}$ such that $$\omega_t(u_i(t),u_j(t))=\omega_t(v_i(t),v_j(t))=0,\ \ \ \ \ \ \ \omega_t(u_i(t),v_j(t))=\delta_{ij}$$ for all $t\in\mathbb R$? It's possible to construct $S(t)$ in a small neighborhood of every $t_0\in\mathbb R$, but I can't do this for all $t$. Thanks for helping.

Yes, we can do that.
Let $S$ be the set consists of all non-degenerated skew symmetric matrices in $Mn(\mathbb{R},2n)$. And define $p:Gl(\mathbb{R},2n)\to S$ by $p(X)=X^{T}JX$, where J is the following matrix:
\begin{pmatrix}
0 & -I_n\\
I_n & 0\\
\end{pmatrix}
then by Proposition 4.3 of Helgason's "Differential Geometry, Lie groups, and Symmetric Spaces", we have a diffeomorphism $\tilde{p}:Sp(\mathbb{R},2n)\backslash Gl(\mathbb{R},2n) \to S$.
Then your question is reformulated as lifting a smooth curve in $S$ to $Gl(\mathbb{R},2n)$. And we find it's trivial to do it locally by the structure of homogeneous space. Then use Lebesgue method on the integral where the curve is defined, we can glue them up.
