The closed set det(A - I) = 0 has empty interior

Closed. $p(A)=\det(A-I)$ is a polynomial in the entries, hence continuous, and $Z=p^{-1}(\{0\})$ is closed.

Empty interior. $p$ is not the zero polynomial: $p(0)=\det(-I)=(-1)^n\ne0$. By the first problem of this section, $\{p=0\}$ has empty interior.

Direct argument. If $A\in Z$ then $A+\varepsilon I$ for small $\varepsilon\neq0$ is not in $Z$: $1$ is an eigenvalue of $A$, so the eigenvalues of $A+\varepsilon I$ are those of $A$ shifted by $\varepsilon$, and $1$ is an eigenvalue of $A+\varepsilon I$ iff $1-\varepsilon$ is an eigenvalue of $A$. Since $A$ has finitely many eigenvalues, this fails for all small $\varepsilon\ne0$.

Remark. Same argument for any $\lambda\in\mathbb{R}$: $\{\det(A-\lambda I)=0\}$ is a closed hypersurface with empty interior. Lesson. Zero sets of nonzero polynomials are "thin".