I proved that the determinant of a rotational matrix(which represents a rotation in 3D) must be $+1$ or $-1$, but I am not sure why it can't be $-1$.
I proved it using the fact that the Matrix must be orthogonal, therefore the determinant squared must be $1$.
You can perform a rotation "in time". Let $T$ be the rotation by some angle $\phi>0$ around the axis ${\bf a}$. Then there is a continuous map $$f:\quad[0,\phi]\to O(3),\qquad t\mapsto A(t)$$ with $A(0)={\rm id}$ and $A(\phi)=T$, namely $A(t)$ is the rotation by the angle $t$ around ${\bf a}$. Since ${\rm det}\bigl(A(0)\bigr)=1$ and the determinant is a continuous function on the space of linear maps it follows that ${\rm det}(T)={\rm det}\bigl(A(\phi)\bigr)=1$.