Matrices with a real eigenvalue: closed set, interior for n = 2

Idea. Closedness is a compactness argument (eigenvalues cannot escape to infinity). For $n=2$ the condition becomes a single inequality on $(\operatorname{tr},\det)$.

Step 1: $\mathcal R$ is closed. Let $A_k\to A$ with $A_k\in\mathcal R$ and real eigenvalues $\lambda_k$. If $A_kv=\lambda_kv$ with $|v|=1$, then $|\lambda_k|\le\|A_k\|$. As $(A_k)$ converges, $\|A_k\|$ is bounded, so $(\lambda_k)$ is bounded and has a convergent subsequence $\lambda_{k_j}\to\lambda\in\mathbb{R}$. By continuity of $\det$,
$$\det(A-\lambda I)=\lim_j\det(A_{k_j}-\lambda_{k_j}I)=0,$$
so $\lambda$ is a real eigenvalue of $A$ and $A\in\mathcal R$.

Step 2: the case $n=2$. The characteristic polynomial is $t^2-(\operatorname{tr}A)t+\det A$, with discriminant $\Delta(A)=(\operatorname{tr}A)^2-4\det A$. It has a real root iff $\Delta\ge0$:
$$\mathcal R=\{\Delta\ge0\}.$$

Step 3: interior. We claim $\operatorname{int}\mathcal R=\{\Delta>0\}$.
- $\{\Delta>0\}$ is open (as $\Delta$ is continuous) and contained in $\mathcal R$.
- Let $\Delta(A)=0$ with $A=\begin{pmatrix}a&b\\c&d\end{pmatrix}$ and let $J=\begin{pmatrix}0&-1\\1&0\end{pmatrix}$. Then $A+\varepsilon J=\begin{pmatrix}a&b-\varepsilon\\c+\varepsilon&d\end{pmatrix}$ has the same trace and $\det(A+\varepsilon J)=\det A-\varepsilon(b-c)+\varepsilon^2$, so
$$\Delta(A+\varepsilon J)=\Delta(A)+4\varepsilon(b-c)-4\varepsilon^2=4\varepsilon(b-c)-4\varepsilon^2.$$
If $b=c$ this is $-4\varepsilon^2<0$ for any $\varepsilon\neq0$; if $b\neq c$ choose the sign of $\varepsilon$ so that $\varepsilon(b-c)<0$. In both cases $\Delta(A+\varepsilon J)<0$ for small $\varepsilon\ne0$, so $A+\varepsilon J\notin\mathcal R$ and $A\notin\operatorname{int}\mathcal R$.

Check. $A=\begin{pmatrix}1&1\\0&1\end{pmatrix}$ ($\Delta=0$). For $\varepsilon=-0.01$ the eigenvalues of $A+\varepsilon J$ are $1\pm0.1005\,i$ (non-real), while for $\varepsilon=+0.01$ they are $1\pm0.0995$ (real). So $A$ is a boundary point.

Remark.
- For odd $n$ the characteristic polynomial has odd degree, so every real matrix has a real eigenvalue and $\mathcal R=M_n(\mathbb{R})$ trivially.
- For $n=2$ the boundary of $\mathcal R$ is $\{\Delta=0\}$: matrices with a double real eigenvalue. These are the places where a pair of real eigenvalues collides and leaves the real line.