The square roots of A in M_2(R) form two disjoint planes

Idea. Fixing the trace and the determinant, the only freedom left is $w\in\mathbb{C}$, together with a sign for $\operatorname{Im}z$.

Step 1: trace and determinant. Let $A^2=-I$. The minimal polynomial of $A$ divides $t^2+1$, which is irreducible over $\mathbb{R}$, so it equals $t^2+1$. Since it has degree $2=n$, it is the characteristic polynomial: $\operatorname{tr}A=0$ and $\det A=1$. Conversely, if $\operatorname{tr}A=0$ and $\det A=1$, Cayley-Hamilton gives $A^2-0\cdot A+I=0$, i.e. $A^2=-I$. Hence
$$\{A^2=-I\}=\{\operatorname{tr}A=0,\ \det A=1\}.$$

Step 2: coordinates. The conditions read $\operatorname{Re}z=0$ and $|z|^2-|w|^2=1$. Writing $z=iq$ with $q\in\mathbb{R}$ we get $q^2=1+|w|^2$, that is
$$q=+\sqrt{1+|w|^2}\quad\text{or}\quad q=-\sqrt{1+|w|^2}.$$
Note $q\ne0$ since $q^2\ge1$. So the set is the union of two graphs over $w\in\mathbb{C}$:
$$S_\pm=\{(z,w):z=\pm i\sqrt{1+|w|^2}\}.$$

Step 3: each sheet is a plane. The projection $A\mapsto w$ is continuous on $S_+$, and it has the continuous inverse $w\mapsto(i\sqrt{1+|w|^2},w)$ (then back to the matrix by the inverse linear map). So $S_+\cong\mathbb{C}\cong\mathbb{R}^2$, and likewise $S_-$.

Step 4: the sheets are separated. $S_+$ and $S_-$ are disjoint (their $\operatorname{Im}z$ have opposite signs and $|\operatorname{Im}z|\ge1$). Each is closed (the zero set of a continuous equation inside the set), and the set is their union, so each is also open in the union (its complement there is the other, closed, sheet). Therefore the topological space is $S_+\sqcup S_-\cong\mathbb{R}^2\sqcup\mathbb{R}^2$.

Check. $J=\begin{pmatrix}0&-1\\1&0\end{pmatrix}$ has $z=i$, $w=0$: it sits at the bottom of the sheet $S_+$ (rotation by $+90^\circ$). The matrix $-J$ has $z=-i$, $w=0$ and lies on $S_-$.

Remark. The two sheets are exchanged by $A\mapsto-A$, and $\operatorname{sign}(\operatorname{Im}z)=\operatorname{sign}(c-b)$ records the orientation in which $A$ "rotates" the plane. This is the case $n=2$ of the theorem that $\{A\in M_{2m}(\mathbb{R}):A^2=-I\}$ has exactly two components (see the problem on complex structures).

Lesson. Slice the coordinate space by the equations given by trace and determinant, solve for one coordinate, and read the topology from the resulting graphs.