= Solution
Let $V=\mathbb R^4$ have <symplectic basis> $(e_1,f_1,e_2,f_2)$. Fix $\mathrm{vol}=e_1\wedge f_1\wedge e_2\wedge f_2$ and define a symmetric <bilinear form> $Q$ on $\Lambda^2V$ by $u\wedge v=Q(u,v)\mathrm{vol}$. The pairings between complementary basis bivectors split this six-dimensional space into three hyperbolic planes, so $Q$ has signature $(3,3)$.
The <symplectic group> preserves volume and the bivector $\Omega=e_1\wedge f_1+e_2\wedge f_2$, obtained from the inverse symplectic form, up to its fixed sign convention. Since $Q(\Omega,\Omega)=2$, its orthogonal complement $W=\Omega^\perp$ has dimension five and signature $(2,3)$. On $W$ use $-Q$, which has signature $(3,2)$. This constructs the <exterior-square double cover of SO0(3,2)>.
If an element acts trivially on $W$, it also fixes $\Omega$, so its action on the entire <exterior square> is the identity. Every decomposable bivector then shows that its two-plane is preserved. A line is the intersection of two such planes; hence every line is preserved, and the matrix is scalar. Its exterior-square action forces its scalar square to equal one, giving precisely $\pm I$.
The derivative is injective by the same infinitesimal argument: zero action on $\Lambda^2V$ forces an infinitesimal scalar, whose induced scalar there is twice its value and therefore zero. The <symplectic Lie algebra> has dimension $2^2+2(2\cdot3/2)=10$, from blocks $\left(\begin{smallmatrix}A&B\\C&-A^T\end{smallmatrix}\right)$ with symmetric $B,C$. The orthogonal Lie algebra in five dimensions also has dimension ten. Thus the image is open. Real symplectic polar decomposition has connected compact factor $U(2)$ and a contractible positive symplectic factor, so $Sp(4,\mathbb R)$ is connected and its image is exactly the identity component:
$$
\boxed{Sp(4,\mathbb R)/\{\pm I\}\cong SO_0(3,2).}
$$
As in the preceding noncompact cases, this does not identify the quotient with the disconnected full $SO(3,2)$.
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