To show that 1D potentials have no degenerate bound states, Griffiths and Schroeter, Introduction to Quantum Mechanics (3rd Edition) suggest considering two wavefunctions $\psi_1 ( x ) $ and $\psi_2 ( x ) $ that satisfy the Schrödinger equation with the same energy $$ -\frac{\hbar^2}{2 m} \frac{d^2}{d x^2} \psi_1 (x) + V (x) \psi_1 (x) = E \psi_1 (x), \;\;\;(1),$$ $$ -\frac{\hbar^2}{2 m} \frac{d^2}{d x^2} \psi_2 (x) + V (x) \psi_2 (x) = E \psi_2 (x), \;\;\;(2).$$ Multiplying (1) by $\psi_2$ and (2) by $\psi_1$ and subtracting gives $$ -\frac{\hbar^2}{2 m} \left( \psi_2 \frac{d^2}{d x^2} \psi_1 - \psi_1 \frac{d^2}{d x^2} \psi_2 \right) = 0 $$ from which it is easily seen is the same as $$ \frac{d }{d x} \left( \psi_2 \frac{d}{d x} \psi_1 - \psi_1 \frac{d}{d x} \psi_2 \right) = 0$$ that implies $\psi_2 \frac{d}{d x} \psi_1 - \psi_1 \frac{d}{d x} \psi_2 = $ constant. Using the boundedness of the wave functions, the constant is zero and leads to the result $ \psi_1, \psi_2$ are linearly dependent.
But can't you use the same argument for the 3D Schrödinger equation? In detail
$$ -\frac{\hbar^2}{2 m} \nabla^2 \psi_1 ( {\bf r} ) + V ({\bf r}) \psi_1 ({\bf r}) = E \psi_1 ({\bf r}), \;\;\;(3),$$ $$ -\frac{\hbar^2}{2 m} \nabla^2 \psi_2 ({\bf r}) + V ({\bf r}) \psi_2 ({\bf r}) = E \psi_2 ({\bf r}), \;\;\;(4).$$ Playing the same game as for the 1D case gives $$ \psi_2 \nabla^2 \psi_1 - \psi_1 \nabla^2 \psi_2 = 0 \;\;\;(5) $$ that can be written as $$ \nabla {\large \cdot} \left( \psi_2 \nabla \psi_1 - \psi_1 \nabla \psi_2 \right) = 0 \;\;\;(6) $$
It seems then that (6) $\implies \psi_2 \nabla \psi_1 - \psi_1 \nabla \psi_2 = $ constant (vector) that can be set to zero because of the boundedness of the solutions, again leading to the conclusion $ \psi_1, \psi_2$ are linearly dependent. But, it is known that the 3D Schrödinger equation may have degenerate solutions, so what's gone wrong?
