Let $M_n$ be the set of $n\times n$ real matrices, identified with $L(\mathbb{R}^n,\mathbb{R}^n)$ in the usual way. Let $U \subset M_n$ be the subset $$U = \{ X \in M_n | I-X \text{ is invertible} \}$$
Show that $U$ is an open subset of $M_n$ and it contains $V = \{X | \ ||X|| < 1\}$ where $||\cdot||$ is the operator norm.
So far the only way I have managed to come up with is to show that $M_n$ is complete, then "Guess" the inverse of $I-X$ as $I+X+X^2+\cdots$ then show that this converges when $||X|| < 1$ so $V \subset U$. Then use this to prove that the set of invertible matrices is open and then apply the homeomorphism $f(X) = I-X$ to show $U$ is open.
As you can see that is very complicated, this question is just one small section of a longer question designed to be done in 30 minutes so I am probably missing a much easier way of solving the problem. Any help is much appreciated.
Sketch:
If we write $$ X=\begin{bmatrix}x_{11}&\dots&x_{1n}\\\vdots&\ddots&\vdots\\x_{n1}&\dots&x_{nn}\end{bmatrix}, $$ then $$ \det(I-X)=\det\begin{bmatrix}1-x_{11}&\dots&-x_{1n}\\\vdots&\ddots&\vdots\\-x_{n1}&\dots&1-x_{nn}\end{bmatrix} $$ is a polynomial in the $x_{ij}$'s (use the cofactor/symmetric group expansion) so the determinant is continuous in the entries of the matrix $X$. Hence the preimage of $\mathbb{R}\setminus\{0\}$ is an open set (and the open set is $U$).
For the second part, note that $I-X$ is not invertible is the same as $X$ has an eigenvalue of $1$. The condition that $\|X\|<1$ implies that all eigenvalues of $X$ have norm less than $1$.