 Solution

By continuity on the compact closure, $v$ has a maximum and minimum. If its maximum were greater than one, the boundary hypothesis would force that maximum to occur at an interior point $x_0$. There $\Delta v(x_0)\le 0$, whereasa contradiction. If its minimum were less than minus one, it would again be interior, where $\Delta v\ge 0$ but $v^3-v<0$. ThereforeThis is a direct maximum/minimum argument for a nonlinear elliptic boundary value problem; treating its nonlinearity as a zeroth-order coefficient of fixed nonpositive sign would not be justified.