Level sets of the determinant

Idea. All the non-zero level sets are copies of $SL_n$ obtained by a scaling and perhaps a reflection, hence they are all homeomorphic to $SL_n(\mathbb{R})$.

Case $c>0$. Let $s=c^{1/n}>0$ and define $\Phi(A)=sA$. Since $\det(sA)=s^n\det A=c\det A$, $\Phi$ maps $SL_n$ into $\det^{-1}(c)$. It is a bijection onto it: its inverse is $B\mapsto s^{-1}B$, because $\det(s^{-1}B)=c^{-1}\cdot c=1$. Both maps are continuous, so $\det^{-1}(c)\cong SL_n(\mathbb{R})$.

Case $c<0$. Let $D=\operatorname{diag}(-1,1,\dots,1)$ and $s=|c|^{1/n}$. Define $\Psi(A)=sDA$. Then $\det(sDA)=s^n\cdot(-1)\cdot\det A=-|c|=c$ for $A\in SL_n$. Again $\Psi$ is continuous with continuous inverse $B\mapsto s^{-1}DB$ (note $D^{-1}=D$), so $\det^{-1}(c)\cong SL_n(\mathbb{R})$.

Conclusion. $\det^{-1}(c)$ is the continuous image of the path connected set $SL_n(\mathbb{R})$, hence path connected.

Remark. For $c=0$ the level set is the set of singular matrices, which is also path connected (Problem 3), but for a very different reason (it is a cone, not a group). The hypothesis $n\ge2$ is not even necessary here: for $n=1$ each level set is a single point.