quantum mechanics - Time-independent Schrödinger function: If the potential $V$ is even, then the wave function $psi$ can always be taken to be either even or odd

I have done the Problem 2.1 in Griffiths' quantum mechanics, and it seems not making sense to me.

What if the wave function isn't symmetric at all? Then obviously the proof doesn't work. The solution confuses me.

If $V(x)=V(-x)$ then the "position" wave function can always be taken to be either even or odd.