Continuity of the inverse matrix function

For a differentiation module I am taking one of the exercises (not homework) asks:

Show that the set $U \subset \mathbb{R}^{n^{2}}$ of matrices $A$ with $det(A) \neq 0$ is open. Let $A^{-1}$ be the inverse of the matrix $A$. Show that the mapping $A \mapsto A^{-1}$ is continuous from $U$ to $U$.

This lecturer has been known to make mistakes in his exercise sheets and I am not sure if the question is even answerable since we have not been given a way to define a neighborhood around a matrix... but here is what I have said so far:

My solution to the first part is that $det(A)$ can be expressed as a polynomial in the entries of $A$, and since polynomial functions are continuous we have that the determinant function is continuous from which we can say the given set is indeed open. Any thoughts on the next part?

First: $f=\det$ is continuous, and the set $\operatorname{GL}_n(\mathbb{R})$ of all invertible matrices can be written as $$ \operatorname{GL}_n(\mathbb{R}) = f^{-1}(\mathbb{R}^\ast) $$ Since $\mathbb{R}^\ast$ is open and $f$ is continuous, $f^{-1}(\mathbb{R}^\ast)$ is open.

Now, for showing $g\colon A\in \operatorname{GL}_n(\mathbb{R}) \mapsto A^{-1}$ is continuous, use the fact that $\det$ is continuous, that $A\mapsto\operatorname{adj} A$ is continuous (not hard to show), and that $$ A^{-1} = \frac{1}{\det A}\operatorname{adj} A. $$

You're on the right track. You have been given a way to define a neighborhood. He's told you in the problem to think of $n\times n$ matrices as $\Bbb R^{n^2}$.

HINT: Do you have a concrete formula for the inverse of a nonsingular matrix? Say, something involving cofactors?