The prohibition against degeneracy of the energy levels in one-dimension is given in standard textbooks.¹

Assume that there are two solutions ${\psi }_a$ and ${\psi }_b$ that have the same energy $E$. Each will satsify the Time-Independent Schrodinger equation as follows

$$\begin{array}{}\text{(1)}& -\frac{{\hslash }^2}{2m}\frac{d^2{\psi }_a}{d{x}^2}+V(x){\psi }_a=E{\psi }_a\end{array}$$

$$\begin{array}{}\text{(2)}& -\frac{{\hslash }^2}{2m}\frac{d^2{\psi }_b}{d{x}^2}+V(x){\psi }_b=E{\psi }_b\end{array}$$

Multiplying Eq.$\text{(1)}$ by ${\psi }_b$ and multiplying Eq.$\text{(2)}$ by ${\psi }_a$ results in

$$\begin{array}{}\text{(3)}& -\frac{{\hslash }^2}{2m}(\frac{d^2{\psi }_a}{d{x}^2}){\psi }_b+V(x){\psi }_a{\psi }_b=E{\psi }_a{\psi }_b\end{array}$$

$$\begin{array}{}\text{(4)}& -\frac{{\hslash }^2}{2m}(\frac{d^2{\psi }_b}{d{x}^2}){\psi }_a+V(x){\psi }_b{\psi }_a=E{\psi }_b{\psi }_a\end{array}$$

Subtracting Eq.$\text{(4)}$ from Eq.$\text{(3)}$ yields

$$\begin{array}{}\text{(5)}& -\frac{{\hslash }^2}{2m}[(\frac{d^2{\psi }_a}{d{x}^2}){\psi }_b–(\frac{d^2{\psi }_b}{d{x}^2}){\psi }_a]=0\end{array}$$

Equation $\text{(5)}$ can be rewritten as

$$\begin{array}{}& -\frac{{\hslash }^2}{2m}\cdot \frac{d}{dx}[{\psi }_b(\frac{d{\psi }_a}{dx})–{\psi }_a(\frac{d{\psi }_b}{dx})]=0\end{array}$$

Which implies that

$$\begin{array}{}\text{(6)}& {\psi }_b(\frac{d{\psi }_a}{dx})–{\psi }_a(\frac{d{\psi }_b}{dx})=CC=\text{constant}\end{array}$$

Since this relationship holds true at any point $x$, then taking the standard assumption that

$$\begin{array}{}& \underset{x\to \pm \mathrm{\infty }}{lim}{\psi }_a(x)=0\end{array}$$

$$\begin{array}{}& \text{and}\end{array}$$

$$\begin{array}{}& \underset{x\to \pm \mathrm{\infty }}{lim}{\psi }_b(x)=0\end{array}$$

results in $C=0$ which should be constant for $-\mathrm{\infty }<x<+\mathrm{\infty }$.

Setting $C=0$, rearranging Eq.$\text{(6)}$ gives the following relationship

$$\begin{array}{}\text{(7)}& \frac{1}{{\psi }_a}(\frac{d{\psi }_a}{dx})=\frac{1}{{\psi }_b}(\frac{d{\psi }_b}{dx})\end{array}$$

resulting in

$$\begin{array}{}& \text{ln}({\psi }_a)=\text{ ln}({\psi }_b)+cc=\text{constant}\end{array}$$

as the relationship between ${\psi }_a$ and ${\psi }_b$. Expressing the relationship in exponential form gives

$$\begin{array}{}& {\psi }_a=e^c{\psi }_b\end{array}$$

showing that ${\psi }_a$ and ${\psi }_b$ are not linearly independent functions.

Note that in order to proceed from Eq.$\text{(6)}$ to Eq.$\text{(7)}$, it is assumed that the first derivative is continuous everywhere.

For the particle in an infinite well, the assumption of continuity of the first derivative $d\psi /dx$ is false. Even in the traditional approach in which the value of the wave function $\psi$ is equated with zero at the boundary, it is accepted that the first derivative $d\psi /dx$ is not continuous at the boundary. Therefore, for the particle in a box, the prohibition against degeneracy is not applicable.