[proofplan]
We prove the basis statement by induction on $\dim V$. Choose a nonzero vector $e_1$, use nondegeneracy of $\omega$ to find $f_1$ with $\omega(e_1,f_1)=1$, and split off the symplectic plane $S=\operatorname{span}(e_1,f_1)$. The $\omega$-orthogonal complement $S^\omega$ is again symplectic, so induction gives a symplectic basis there. Concatenating the two bases gives a symplectic basis of $V$, and the corresponding coordinate map from $\mathbb R^{2n}$ identifies $\omega$ with the standard form $\omega_0$.
[/proofplan]
[step:Start the induction and split off a symplectic plane]
We prove the result by strong induction on the integer $m=\dim V$.
If $m=0$, then the empty ordered list is a basis of $V=\{0\}$ and satisfies the required pairing relations vacuously. This is the case $n=0$.
Assume now that $m>0$ and that the theorem holds for every real symplectic [vector space](/page/Vector%20Space) of dimension strictly smaller than $m$. Choose a nonzero vector $e_1\in V$. Since $\omega$ is nondegenerate, the linear functional
\begin{align*}
\omega(e_1,\cdot):V\to\mathbb R
\end{align*}
is not the zero functional. Hence there exists a vector $g\in V$ such that $\omega(e_1,g)\ne 0$. Define
\begin{align*}
f_1:=\frac{1}{\omega(e_1,g)}g.
\end{align*}
By bilinearity of $\omega$,
\begin{align*}
\omega(e_1,f_1)=1.
\end{align*}
Let $S\subset V$ be the two-dimensional subspace
\begin{align*}
S:=\operatorname{span}\{e_1,f_1\}.
\end{align*}
The vectors $e_1$ and $f_1$ are linearly independent: if $f_1=ae_1$ for some $a\in\mathbb R$, then skew-symmetry gives
\begin{align*}
1=\omega(e_1,f_1)=a\omega(e_1,e_1)=0,
\end{align*}
a contradiction. Thus $\dim S=2$.
[guided]
We begin by finding the smallest possible symplectic building block: a two-dimensional plane on which $\omega$ is already nondegenerate. Pick a nonzero vector $e_1\in V$. Nondegeneracy of $\omega$ means precisely that no nonzero vector pairs to zero with every vector in $V$. Therefore the map
\begin{align*}
\omega(e_1,\cdot):V\to\mathbb R
\end{align*}
is a nonzero linear functional. Since it is nonzero, there is some vector $g\in V$ with $\omega(e_1,g)\ne 0$.
We normalize $g$ so that the pairing is exactly $1$. Define
\begin{align*}
f_1:=\frac{1}{\omega(e_1,g)}g.
\end{align*}
Bilinearity gives
\begin{align*}
\omega(e_1,f_1)=\omega\left(e_1,\frac{1}{\omega(e_1,g)}g\right)=\frac{1}{\omega(e_1,g)}\omega(e_1,g)=1.
\end{align*}
Now define the candidate symplectic plane
\begin{align*}
S:=\operatorname{span}\{e_1,f_1\}\subset V.
\end{align*}
The vectors $e_1$ and $f_1$ must be linearly independent. Indeed, if $f_1=ae_1$ for some scalar $a\in\mathbb R$, then skew-symmetry of $\omega$ gives $\omega(e_1,e_1)=0$, so
\begin{align*}
1=\omega(e_1,f_1)=\omega(e_1,ae_1)=a\omega(e_1,e_1)=0,
\end{align*}
which is impossible. Hence $\dim S=2$.
[/guided]
[/step]
[step:Show the complement of the symplectic plane is symplectic]
Define the $\omega$-orthogonal complement of $S$ by
\begin{align*}
S^\omega:=\{v\in V:\omega(v,s)=0\text{ for every }s\in S\}.
\end{align*}
First, the restriction $\omega|_{S\times S}$ is nondegenerate. Indeed, if $u=ae_1+bf_1\in S$ and $\omega(u,s)=0$ for every $s\in S$, then applying this to $s=e_1$ gives
\begin{align*}
0=\omega(u,e_1)=a\omega(e_1,e_1)+b\omega(f_1,e_1)=-b,
\end{align*}
so $b=0$. Applying it to $s=f_1$ gives
\begin{align*}
0=\omega(u,f_1)=a\omega(e_1,f_1)+b\omega(f_1,f_1)=a,
\end{align*}
so $a=0$. Hence $u=0$.
Let $S^*$ denote the dual vector space of $S$, that is, the vector space of linear maps $S\to\mathbb R$. Define the [linear map](/page/Linear%20Map) $\rho:V\to S^*$ by
\begin{align*}
\rho(v)=\omega(v,\cdot)|_S.
\end{align*}
Its kernel is exactly $S^\omega$. Since $\omega|_{S\times S}$ is nondegenerate, the restricted map $\rho|_S:S\to S^*$ is an isomorphism. Therefore $\rho$ is surjective, so the [rank-nullity theorem](/theorems/916) for finite-dimensional linear maps gives
\begin{align*}
\dim S^\omega=\dim V-\dim S^*=\dim V-2.
\end{align*}
Also $S\cap S^\omega=\{0\}$ because $\omega|_{S\times S}$ is nondegenerate. Since
\begin{align*}
\dim S+\dim S^\omega=2+(\dim V-2)=\dim V,
\end{align*}
we have the [direct sum](/page/Direct%20Sum) decomposition
\begin{align*}
V=S\oplus S^\omega.
\end{align*}
It remains to check that $\omega|_{S^\omega\times S^\omega}$ is nondegenerate. Let $v\in S^\omega$ satisfy $\omega(v,w)=0$ for every $w\in S^\omega$. Since $v\in S^\omega$, also $\omega(v,s)=0$ for every $s\in S$. The decomposition $V=S\oplus S^\omega$ then implies $\omega(v,z)=0$ for every $z\in V$. By nondegeneracy of $\omega$ on $V$, we get $v=0$. Thus $(S^\omega,\omega|_{S^\omega\times S^\omega})$ is a symplectic vector space.
[/step]
[step:Apply induction to the symplectic complement and concatenate bases]
The vector space $S^\omega$ has dimension $\dim V-2<m$, so the induction hypothesis applies to the symplectic vector space $(S^\omega,\omega|_{S^\omega\times S^\omega})$. Therefore there are vectors
\begin{align*}
e_2,\dots,e_n,f_2,\dots,f_n\in S^\omega
\end{align*}
forming a basis of $S^\omega$ and satisfying
\begin{align*}
\omega(e_i,e_j)=0,\qquad \omega(f_i,f_j)=0,\qquad \omega(e_i,f_j)=\delta_{ij}
\end{align*}
for all $2\le i,j\le n$.
Because $V=S\oplus S^\omega$ and $(e_1,f_1)$ is a basis of $S$, the ordered list
\begin{align*}
(e_1,\dots,e_n,f_1,\dots,f_n)
\end{align*}
is a basis of $V$. For $i\ge 2$, the vectors $e_i$ and $f_i$ lie in $S^\omega$, so $\omega(e_i,e_1)=\omega(e_i,f_1)=\omega(f_i,e_1)=\omega(f_i,f_1)=0$. Skew-symmetry then gives the corresponding vanishings with $e_1$ or $f_1$ in the first argument. Together with $\omega(e_1,f_1)=1$, $\omega(e_1,e_1)=0$, $\omega(f_1,f_1)=0$, and the induction relations on $S^\omega$, this gives
\begin{align*}
\omega(e_i,e_j)=0
\end{align*}
\begin{align*}
\omega(f_i,f_j)=0
\end{align*}
\begin{align*}
\omega(e_i,f_j)=\delta_{ij}
\end{align*}
for all $1\le i,j\le n$.
Thus $V$ admits a symplectic basis. In particular, the basis has $2n$ vectors, so $\dim V=2n$ is even.
[/step]
[step:Build the linear symplectomorphism from the symplectic basis]
Let $(E_1,\dots,E_n,F_1,\dots,F_n)$ denote the standard ordered basis of $\mathbb R^{2n}$, where $E_i$ is the coordinate vector in the $q_i$-direction and $F_i$ is the coordinate vector in the $p_i$-direction. For the standard symplectic form
\begin{align*}
\omega_0=\sum_{i=1}^n dq_i\wedge dp_i,
\end{align*}
we have
\begin{align*}
\omega_0(E_i,E_j)=0
\end{align*}
\begin{align*}
\omega_0(F_i,F_j)=0
\end{align*}
\begin{align*}
\omega_0(E_i,F_j)=\delta_{ij}.
\end{align*}
Define the linear map $\Phi:\mathbb R^{2n}\to V$ by requiring
\begin{align*}
\Phi(E_i)=e_i
\end{align*}
and
\begin{align*}
\Phi(F_i)=f_i
\end{align*}
for every $1\le i\le n$.
Since $(e_1,\dots,e_n,f_1,\dots,f_n)$ is a basis of $V$, the map $\Phi$ is a linear isomorphism. To prove that $\Phi$ is symplectic, it suffices by bilinearity to check the equality
\begin{align*}
\Phi^*\omega=\omega_0
\end{align*}
on pairs of standard basis vectors. For all $1\le i,j\le n$,
\begin{align*}
(\Phi^*\omega)(E_i,E_j)=\omega(e_i,e_j)=0=\omega_0(E_i,E_j),
\end{align*}
\begin{align*}
(\Phi^*\omega)(F_i,F_j)=\omega(f_i,f_j)=0=\omega_0(F_i,F_j),
\end{align*}
and
\begin{align*}
(\Phi^*\omega)(E_i,F_j)=\omega(e_i,f_j)=\delta_{ij}=\omega_0(E_i,F_j).
\end{align*}
Skew-symmetry then also gives equality on the reversed pairs $(F_j,E_i)$. Hence $\Phi^*\omega=\omega_0$, so $\Phi$ is a linear symplectomorphism from $(\mathbb R^{2n},\omega_0)$ to $(V,\omega)$.
[/step]
