Proof of Darboux theorem in symplectic geometry using Moser theorem Darboux theorem in symplectic geometry says the following:

Let $(M^{2n},\omega)$ be a symplectic manifold. If $p\in M$, there is a chart $(U,x_1,...,x_n,y_1,...,y_n)$ centered at $p$ such that:
  $$\omega=\sum_{i=1}^ndx_i\wedge dy_i$$

In Ana Cannas' Lectures on Symplectic Geometry, the proof begins like this: 
"Use any symplectic basis for $T_pM$ to construct coordinates $(x'_
1,..., x'_n, y'_1,...,y'_n)$ centered at $p$ and valid on some neighborhood $U_0$, so that $\omega_p=\sum_{i=1}^n(dx'_i\wedge dy'_i)_p$".
Then she uses Moser theorem, which guarantees there is a neighbourhood $U_1$ and a diffeomorphism $\varphi:U_0\to U_1$ with $\varphi(p)=p$ and $\varphi^*\left(\sum_idx'_i\wedge dy'_i\right)=\omega$. Then we only need to define $x_i:=x'_i\circ\varphi$, $y_i:=y'_i\circ\varphi$.
The final part is clear to me, but the beginning is not. If we have a symplectic basis $\{e_1,f_1,...,e_n,f_n\}$ of $T_pM$, how can we construct the coordinates $(x'_1,...,x'_n,y'_1,...,y'_n)$?
 A: Say we have a symplectic basis $\{e_{1},\ldots,e_{n},f_{1},\ldots,f_{n}\}$ for $(T_{p}M,\omega_{p})$. If $\{e_{1}^{*},\ldots,e_{n}^{*},f_{1}^{*},\ldots,f_{n}^{*}\}$ is the dual basis of $\left(T_{p}M\right)^{*}$, then $\omega_{p}\in\wedge^{2}\left(T_{p}M\right)^{*}$ has the following expression:
$$
\omega_{p}=\sum_{i=1}^{n}e_{i}^{*}\wedge f_{i}^{*}.
$$
So we have to show that there exists a coordinate chart $(x_{1}',\ldots,x_{n}',y_{1}',\ldots,y_{n}')$ around $p$ satisfying
$$
(dx_{i}')_{p}=e_i^{*},\\
(dy_{i}')_{p}=f_{i}^{*}.
$$
Such a chart can be constructed as follows. Let $(\xi_{1},\ldots,\xi_n,\eta_1,\ldots,\eta_{n})$ be any coordinate chart around $p$. Now both $\{e_{1}^{*},\ldots,e_{n}^{*},f_{1}^{*},\ldots,f_{n}^{*}\}$ and $\{(d\xi_{1})_{p},\ldots,(d\xi_{n})_{p},(d\eta_{1})_{p},\ldots,(d\eta_{n})_{p}\}$ are bases of $(T_{p}M)^{*}$. Let the following block matrix be the change of bases:
$$
\begin{bmatrix}
A & C\\ B&E \tag{1}
\end{bmatrix},
$$
i.e.
$$
e_{1}^{*}=\sum_{j=1}^{n}a_{j1}(d\xi_{j})_{p}+\sum_{j=1}^{n}b_{j1}(d\eta_{j})_{p},\\
\vdots\\
e_{n}^{*}=\sum_{j=1}^{n}a_{jn}(d\xi_{j})_{p}+\sum_{j=1}^{n}b_{jn}(d\eta_{j})_{p},\\
f_{1}^{*}=\sum_{j=1}^{n}c_{j1}(d\xi_{j})_{p}+\sum_{j=1}^{n}e_{j1}(d\eta_{j})_{p},\\
\vdots\\
f_{n}^{*}=\sum_{j=1}^{n}c_{jn}(d\xi_{j})_{p}+\sum_{j=1}^{n}e_{jn}(d\eta_{j})_{p}.
$$
This suggests that we should define
$$
x_{1}'=\sum_{j=1}^{n}a_{j1}\xi_{j}+\sum_{j=1}^{n}b_{j1}\eta_{j},\\
\vdots\\
x_{n}'=\sum_{j=1}^{n}a_{jn}\xi_{j}+\sum_{j=1}^{n}b_{jn}\eta_{j},\\
y_{1}'=\sum_{j=1}^{n}c_{j1}\xi_{j}+\sum_{j=1}^{n}e_{j1}\eta_{j},\\
\vdots\\
y_{n}'=\sum_{j=1}^{n}c_{jn}\xi_{j}+\sum_{j=1}^{n}e_{jn}\eta_{j}.
$$
To check that $(x_{1}',\ldots,x_{n}',y_{1}',\ldots,y_{n}')$ are really coordinates around $p$, we should check that the map $$(\xi_{1},\ldots,\xi_n,\eta_1,\ldots,\eta_{n})\mapsto(x_{1}'(\xi,\eta),\ldots,x_{n}'(\xi,\eta),y_{1}'(\xi,\eta),\ldots,y_{n}'(\xi,\eta))$$
is a local diffeomorphism around $p$. This is the case, since its Jacobian is exactly the transpose of the block matrix (1) above, which is non-singular.
