As I understand it, the symplectic Lie group $Sp(2n,\mathbb{R})$ of $2n \times 2n$ symplectic matrices is generated by the matrices in http://en.wikipedia.org/wiki/Symplectic_group#Infinitesimal_generators .
Does this mean that $sl(n,\mathbb{R})$ is a subalgebra of the corresponding lie algebra, since in that formula we can truncate by removing the matrices $B$ and $C$ and enforce that $A$ is traceless?
Also, $sp(4,\mathbb{R})$ has dimension $10$ and $sl(3,\mathbb{R})$ has dimension $8$. Is $sl(3,\mathbb{R})$ a subalgebra of $sp(4,\mathbb{R})$ or not?
As a more practical question, where should I look if I want to look up these kind of standard results on standard Lie groups? Googling did not take me very far.
You have it backwards, in fact $\mathrm{Sp}(2n, \mathbb R) \subseteq \mathrm{SL}(2n, \mathbb R)$ (see here) so $\mathfrak{sp}(2n, \mathbb R) \subseteq \mathfrak{sl}(2n, \mathbb R)$.
You can see that each of those infinitesimal generators has trace $0$ (you don't need $A$ to be traceless for this to be true), but not every trace $0$ matrix must be written in that form.