I tried to prove the simplicity of $G=Sp_{2n}(K)$ ($n\geqslant 2$), where $K$ is any field. If $|K|\geq 4$, then this is fairly easy to prove. To start, we use the following result that is also used to prove simplicity of the other classical groups.
Let $G$ be a group acting on a set $S$, and suppose that the action is primitive and faithful. In addition, suppose that:
a) There exists an $x\in S$ and an abelian normal divisor $H\lhd G_x$ such that $G$ is generated by the conjugates of $H$.
b) $[G,G]=G$
Then $G$ is simple.
Now let $G$ act on $\mathbb{P}^{2n-1}(K)$. This is faithful and primitive. Now choose a basis $e_1,...,e_{2n}\in K^{2n}$ so that $\langle e_1,e_2\rangle=1$ and $Span(e_1,e_2)\perp Span(e_3,...,e_{2n})$. Then we can pick a normal abelian $H\lhd G_{\langle e_1\rangle}$ consisting of matrices of the form: $$\begin{pmatrix} J_{a}&0\\0&I_{2n-2}\end{pmatrix}, a\in K $$ To prove $H$ fits criterium (a), and that $[G,G]=G$, it essentially boils down to proving the following (with induction to $n$):
Let $\tau\in Sp_{2n}(K)$. Then there exist 2-dimensional non-degenerate subspaces $V_1,...,V_k\subseteq K^{2n}$ and morphisms $\tau_i\in Sp(V_i)$ such that $\tau=\tau_1...\tau_k$
Using that $Sp_2(K)=SL_{2}(K)$, it follows easily that $G$ is generated by conjugates of $H$. If $|K|\geqslant 4$, $SL_2(K)$ is simple, and it follows easily that $[G,G]=G$ in that case.
But if $|K|\leqslant 3$, the group $SL_2(K)=Sp_2(K)$ is not simple. So we need another way to prove that $[G,G]=G$ if $n\geqslant 2$. Can you help me out?