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. 

 A: Griffiths, Introduction to QM, uses $\Psi$ and $\psi$ to denote a solution$^1$ to the time-dependent and the time-independent Schrödinger eq., respectively.
For a fixed energy $E$, rather than considering a general solution $\psi$ to the TISE, the book is at this point only interested in finding a generating set $\psi_n$ of solutions, so that a general solution is a linear combination $\psi=\sum_nc_n\psi_n$, cf. the superposition principle.
The purpose of Exercise 2.1.b is to show that it is no loss of generality to assume that the generating element $\psi_n\in\mathbb{R}$ is a real function.
The purpose of Exercise 2.1.c is to show in the case of an even potential $V$ that it is no loss of generality to assume that the generating element $\psi_n$ is an even or an odd function.
The book is not claiming that the general solution $\psi$ has to respect such symmetry.
--
$^1$ Griffiths is implicitly only talking about normalizable solutions in 1D. For unnormalizable solutions the given boundary conditions at $x=\pm\infty$ might violate reality/parity.
