Eigenvalues of $AB$ and $BA$ are the same?

Your proof is correct for $\lambda\neq 0$, because then it isn't possible that you get $Bx=0$ for an eigenvector $x$ of $AB$ to the eigenvalue $\lambda$.

And this is also the general statement: All non zero eigenvalues are the same. That it doesn't works with $\lambda=0$ you see for $A=\begin{pmatrix}1\\0\end{pmatrix}, B=\begin{pmatrix}1&0\end{pmatrix}$.

For another proof look at the characteristic polynomial. See en.wikipedia.org/Characteristic polynomial of a product of two matrices.