Prove there exist a non-zero vector $v$ such that $Av=v$.

If $v\in\mathbb R^3$, then\begin{align}\lVert Av\rVert^2&=\langle Av,Av\rangle\\&=\langle v,A^tAv\rangle\\&=\langle v,v\rangle\text{ (since $A^t=A^{-1}$)}\\&=\lVert v\rVert^2\end{align}and therefore $\lVert Av\rVert=\lVert v\rVert$. On the other hand, the characteristic polynomial of $A$ is a cubic polynomial and therefore it has a real root $\lambda$. So, $\lambda$ is an eigenvalue of $A$; let $v$ be a corresponding eigenvector. We have\begin{align}\lVert v\rVert&=\lVert Av\rVert\\&=\lVert\lambda v\rVert\\&=\lvert\lambda\rvert.\lVert v\rVert\end{align}and therefore $\lambda=\pm1$. If $\lambda=1$, we're done. If $\lambda=-1$, let $W=v^\perp=\{w\in\mathbb R^3\,|\,\langle v,w\rangle=0\}$. Then $A.W\subset W$. The determinant of $A|_W$ must be negative, since $0<\det(A)=(-1)\times\det(A|_W)$. But a $2\times2$ matrix with negative determinant always has eigenvalues (the discriminant of the characteristic polynomial is greater than $0$). Therefore one of the eigenvalue of $A|_W$ is positive and the other one is negative. Since we found a positive eigenvalue, and since we have already proved that the only positive eigenvalue that $A$ can have is $1$, we're done.