= Solution
The function $X\mapsto\det(I-X)$ is a polynomial in the matrix entries, hence <continuous>. Consequently $U$ is the inverse image of the open set $\mathbb R\setminus\{0\}$ and is an <open set>.
If $\|X\|<1$ and $(I-X)v=0$, then $v=Xv$ implies $\|v\|\le\|X\|\|v\|$. This forces $v=0$. The square matrix $I-X$ is therefore <injective>, hence an <invertible matrix>, so
$$
\boxed{V\subseteq U.}
$$
The argument works with any of the induced <operator norms> in part (b).
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