1D Finite potential well: solutions with $\sinh$ and $\cosh$?

Consider the following proof by contradiction:

Suppose you find a state with $E < -V_0$, then its kinetic energy becomes negative at every point $x$ (classically the velocity is imaginary at every point), which means the whole wavefunction (at all $x$) is an evanescent (exponentially decaying) wave, but such a solution is not a physically stable solution (if it exists), i.e. it will not be bounded. Therefore, the original assumption of $E < -V_0$ must have not been right.

Consider the following example: the $V(x)$ you have presented is discontinuous but bounded, therefore, according to the Schrodinger equation the second derivative of $\psi(x)$ is at most discontinuous, therefore the first derivative is continuous, and $\psi(x)$ itself is smooth (on top of being continuous). Now, $\psi(x)$ should be made up of decaying exponentials for all $x$, if you assume $E<-V_0$. You can have different decaying exponentials at different regions but they should be joined in such a way that keeps $\psi(x)$ smooth. But that is not possible: try glueing two decaying (and bounded, of course) exponentials smoothly. The best you will get has a shark spike (not smooth). Alternatively, $\psi(x)$ will be boundless (if forced to be smooth). Now, if you cannot glue together two decaying exponentials (together spanning $-\infty$ to $\infty$), you sure won't be able to glue three. Therefore you need to have $E>-V_0$ to generate an oscillatory region to provide a smooth connection between the two asymptotic decaying parts.