Let $\lambda$ be an eigenvalue of $A$. Prove that $\lambda^{-1}$ is an eigenvalue of $A^{-1}$.

Before I continue, it is important to note that $\lambda$ is a scalar and not a matrix. So given that $A$ is invertible, $Ax=\lambda x$, $A$ is invertible, and $\lambda\neq 0$, we have
$$Ax=\lambda x\implies A^{-1}Ax=A^{-1}\lambda x\implies x=\lambda A^{-1}x\implies \frac1\lambda x=A^{-1}x.$$

Your proof is fine, but you seem to be making things a bit overly complicated.

Proof: Let $x \neq 0$ be such that $Ax = \lambda x$. Since $A$ is invertible, it has a trivial kernel, so $\lambda \neq 0$. We then have $$ Ax = \lambda x \implies\\ A^{-1}Ax = A^{-1}(\lambda x) \implies\\ x = \lambda A^{-1}(x) \implies\\ \frac 1{\lambda} x = A^{-1}x $$ By the above equation, $x$ is an eigenvector of $A^{-1}$ associated with the eigenvalue $1/\lambda$. The conclusion holds.

(As Robert Israel mentioned in comments, you must assume $A$ is invertible).

There is a more direct approach you can take. As you said $$Ax=\lambda x \implies A^{-1}Ax=A^{-1}\lambda x$$ and then $$A^{-1}Ax=A^{-1}\lambda x \implies Ix = A^{-1}(\lambda x) = \lambda A^{-1}x$$ Now dividing both sides by $\lambda$ yields $$\lambda^{-1}x = A^{-1}x$$ i.e. $\lambda^{-1}$ is an eigenvalue of $A^{-1}$.