The set of real normal matrices

1. Closed. $\mathcal{N}$ is the zero set of the continuous (polynomial) map $A\mapsto AA^T-A^TA$.

Contains the families.
- $A$ symmetric: $A^T=A$, so $AA^T=A^2=A^TA$.
- $A$ skew: $A^T=-A$, so $AA^T=-A^2=A^TA$.
- $A$ orthogonal: $AA^T=I=A^TA$.

2. Not a subspace. Take $S=\begin{pmatrix}0&\frac12\\\frac12&0\end{pmatrix}$ (symmetric) and $K=\begin{pmatrix}0&\frac12\\-\frac12&0\end{pmatrix}$ (skew). Both are in $\mathcal{N}$, but
$$S+K=\begin{pmatrix}0&1\\0&0\end{pmatrix}=E_{12},\quad E_{12}E_{12}^T=E_{11},\ \ E_{12}^TE_{12}=E_{22},$$
which differ: $E_{12}\notin\mathcal{N}$. (For $n>2$, add a zero block.)

Not open. $I\in\mathcal{N}$, but $A_\varepsilon=I+\varepsilon E_{12}\to I$ is not normal for $\varepsilon\ne0$:
$$A_\varepsilon A_\varepsilon^T=I+\varepsilon(E_{12}+E_{21})+\varepsilon^2E_{11},\qquad A_\varepsilon^TA_\varepsilon=I+\varepsilon(E_{12}+E_{21})+\varepsilon^2E_{22},$$
and $E_{11}\ne E_{22}$. So no neighbourhood of $I$ lies in $\mathcal{N}$.

Remark. $\mathcal{N}$ is the set of matrices that commute with their transpose; by the spectral theorem, normal real matrices are exactly those orthogonally similar to a block diagonal matrix with $1\times1$ real blocks and $2\times2$ blocks $\begin{pmatrix}a&-b\\b&a\end{pmatrix}$. It is a closed cone (stable under real scaling) but not convex. Lesson. Sums of normal matrices are generally not normal.