I want to prove the following statement.
Let $V$ be a symplectic vector space with a symplectic structure $\omega$, and $W$ be a subspace of $V$ of rank $2r$ and dimension $2r+k$. Then, in suitable Darboux coordinates, $W$ is given by $$p_{r+1}=\dots=p_n=0\text{ and }q_{r+k+1}=\dots=q_n=0.$$
I have a question if I am on the right track with my last step:
First, we can take a symplectic structure $\omega$ and restrict it to $W$. We might have that $\omega$ is degenerate on $W$ if $k>0$ i.e. $\ker(\omega|_{W})\neq\{0\}$. Then consider the quotient space $W/\ker(\omega|_{W})$ and observe that $\omega$ restricted to the quotient is going to be non-degenerate. Indeed, it follows from the fact that if $a,b\in W/\ker(\omega)$, then $a=a_W+a_{\ker}$ and $b=b_W+b_{\ker}$ with $a_{\ker},b_{\ker}\in \ker(\omega)$. So, $$\omega(a,b)=\omega(a_W+a_{\ker},b_W+b_{\ker})=\omega(a_W,b_W)$$ as $\omega$ is a billinear form.
As $\omega$ restricted to $W/\ker(\omega|_{W})$ is a symplectic structure on $W/\ker(\omega|_{W})$, then we can apply Linear Darboux Theorem and obtain coordinates $$\overline{p}_1,\dots,\overline{p}_r,\overline{q}_1,\dots,\overline{q}_r\in W/\ker(\omega|_{W}).$$
Then we can lift $\overline{p}_i$ and $\overline{p}_j$ to $W$ where we choose their kernel part to be zero i.e. we will obtain $$p_1,\dots,p_r,q_1,\dots,q_r\in W$$ with $\omega(p_i,q_j)=\delta_{ij}, \omega(p_i,p_j)=\omega(q_i,q_j)=0$ for $1\leq i,j\leq r$.
Finally, by choosing vectors from $\ker(\omega|_{W})$, we can complete basis $p_1,\dots,p_r,q_1,\dots,q_r$ to a complete Darboux basis $$p_1,\dots,p_n,q_1,\dots,q_n$$ on $V$.
Any comments suggestion to improve/fix the proof will be strongly appreciated.