Matrices with positive definite symmetric part

Openness and convexity. The map $L:A\mapsto A+A^T$ is linear, hence continuous, and $\mathrm{PD}$ is open and convex (it is defined by $x^TSx>0$ for all $x\ne0$, a convex condition; openness by Sylvester's criterion or by compactness of the unit sphere). Therefore $\mathcal{K}=L^{-1}(\mathrm{PD})$ is open and convex.

Eigenvalues. Let $Av=\lambda v$ with $v\in\mathbb{C}^n\setminus\{0\}$. Write $S=\frac{A+A^T}2$, real symmetric, positive definite on $\mathbb{R}^n$, and also on $\mathbb{C}^n$: $v=x+iy$ gives $v^*Sv=x^TSx+y^TSy>0$ (the cross terms cancel because $S$ is symmetric). Since $A$ is real,
$$\overline{v^*Av}=v^TA^T\bar v=(\bar v)^*\ldots$$
more simply, $v^*A^Tv=\overline{v^*Av}$ because $A^T=A^*$ for real $A$. Hence
$$v^*Sv=\tfrac12\big(v^*Av+v^*A^*v\big)=\operatorname{Re}(v^*Av).$$
But $v^*Av=\lambda\,v^*v=\lambda|v|^2$, so
$$\operatorname{Re}\lambda\,|v|^2=v^*Sv>0\ \Longrightarrow\ \operatorname{Re}\lambda>0.$$

Example. $A=\begin{pmatrix}1&5\\-5&1\end{pmatrix}$: $A+A^T=2I>0$ and the eigenvalues are $1\pm5i$, real part $1>0$.

Remark. The converse fails: $\begin{pmatrix}1&10\\0&1\end{pmatrix}$ has both eigenvalues $1$ but $A+A^T=\begin{pmatrix}2&10\\10&2\end{pmatrix}$ is indefinite (determinant $4-100<0$). Positivity of the symmetric part is a *uniform* (for all vectors) version of positivity of the spectrum, and it is related to Lyapunov stability with $P=I$.