lie groups - What's the classification of the algebraic subgroups of Sp(4,R)?

On the one hand, I could not find a published answer with a cursory search. On the other hand, as Ben says, you could work out the answer "by hand". Instead of writing down a sheer list, which might be complicated (and I haven't done the work), I'll write down the main ingredients.

A Zariski-closed subgroup $H$ of any connected semisimple Lie group $G$ has three pieces: (1) finite, (2) connected semisimple, and (3) connected solvable. The Zariski topology forces $H$ to have only finitely many components; if $H_0$ is the connected subgroup, then $H/H_0$ is the finite piece. Then the Lie algebra of $H_0$ has a Levi decomposition, so that you get the other two pieces. The way to analyze the question is to chase down the possibilities for all three pieces.

I think that the finite part always lifts to a slightly larger finite subgroup of $H$. This is not true for groups in general, but I think that it is true in context. Then this finite group is contained in a maximal compact group of $G$. Happily, the compact core of $text{Sp}(4,mathbb{R})$ is $text{SU}(2)$, and the finite subgroups are classified by simply laced Dynkin diagrams.

A semisimple, connected subgroup of $G$ corresponds to a semisimple Lie subalgebra, and that complexifies. The Lie algebra $text{sp}(4,mathbb{C})$ does not have very many inequivalent semisimple subalgebras. From looking a rank, they are isomorphic to $text{sl}(2,mathbb{C})$ or $text{sl}(2,mathbb{C}) oplus text{sl}(2,mathbb{C})$. I am confusing myself a little with the possible positions of the former, although I know there are only a few. The latter embeds in only one way. Then you would work backwards to get the real forms of these complex subalgebras; again there wouldn't be very many.

Finally the solvable part also complexifies and I think that it is contained in a Borel subalgebra at the Lie algebra level.

As for the more general question, for $text{Sp}(2n,mathbb{R})$, there is a tidy converse answer that also shows you that you can't expect a tidy answer for all fixed $n$. Namely, if $G$ is any algebraic group, you can classify its anti-self-dual (or symplectically self-dual) representations. Every algebraic group will have some, because every algebraic group has representations in $text{GL}(n,mathbb{R})$. A more interesting case is when $G$ has an irreducible symplectically self-dual representation. For that purpose, you check that the irreducible representation is real, and then check the Frobenius–Schur indicator.