Real-diagonalisable matrices are not dense

Idea. The set $\mathcal{N}$ of matrices having at least one non-real eigenvalue is open and nonempty, and it is disjoint from the $\mathbb{R}$-diagonalisable matrices (their eigenvalues are real). A set that misses a nonempty open set is not dense.

Nonempty. For $n=2$, the rotation $R=\begin{pmatrix}0&-1\\1&0\end{pmatrix}$ has eigenvalues $\pm i$. For $n\ge2$, the block matrix $R\oplus0_{n-2}$ has a non-real eigenvalue.

Openness (continuity of eigenvalues). Let $A\in\mathcal{N}$ with eigenvalue $\mu$, $\operatorname{Im}\mu=\delta\ne0$. For a matrix $A'$ with eigenvalues $\lambda_1',\dots,\lambda_n'$,
$$\prod_{i=1}^n|\mu-\lambda_i'|=|\chi_{A'}(\mu)|,\qquad\text{so}\qquad\min_i|\mu-\lambda_i'|\le|\chi_{A'}(\mu)|^{1/n}.$$
As $A'\to A$, $\chi_{A'}(\mu)\to\chi_A(\mu)=0$ (the coefficients of $\chi_{A'}$ are continuous in $A'$). So for $A'$ close enough to $A$ there is an eigenvalue $\lambda'$ of $A'$ with $|\lambda'-\mu|<\delta/2$, hence $\operatorname{Im}\lambda'\ne0$: $A'\in\mathcal{N}$.

Conclusion. $\mathcal{N}$ is a nonempty open set containing no diagonalisable matrix, so the real-diagonalisable matrices are not dense.

Remark. In contrast, the matrices diagonalisable over $\mathbb{C}$ are dense (distinct eigenvalues). The discrepancy is that $\mathbb{R}$ is not algebraically closed. Lesson. The inequality $\min_i|\mu-\lambda_i'|\le|\chi_{A'}(\mu)|^{1/n}$ is a cheap way to prove continuity of the spectrum.