Consider a real matrix $B$ defined as
$$ B := X A + A^T X $$
where $X$ is a symmetric positive definite matrix and $A$ has eigenvalues with positive real parts. How can I prove that eigenvalues of $B$ also have positive real parts?
Note that since $B$ is symmetric, its eigenvalues are real.
On
Consider the matrices
$$A=\begin{bmatrix}1 & -1\\ -1 & 3\end{bmatrix}\ \mathrm{and}\ X=\begin{bmatrix}10 & 0\\0 & 1\end{bmatrix}.$$
The matrix $A$ has two real positive eigenvalues. However, we have that $$A^TX+XA=\begin{bmatrix} 20 & -11\\ -11 & 6\end{bmatrix},$$
which has a negative eigenvalues. So, the implication you want to prove it not true.
What can be done is the other way round, though. Assume that $A$ and $B$ are given. If $X$ satisfies this equation, then we have that
$$X=\int_0^\infty e^{-A^Ts}Be^{-As}ds$$
where the integral is well-defined since $A$ has eigenvalues with positive real part.
Indeed, we have that $$\begin{array}{rcl} A^TX+XA&=&\displaystyle A^T\int_0^\infty e^{-A^Ts}Be^{-As}ds+\int_0^\infty e^{-A^Ts}Be^{-As}dsA\\ &=&\displaystyle \int_0^\infty -\dfrac{d}{ds}\left(e^{-A^Ts}Be^{-As}\right)ds\\ &=&B. \end{array}$$
Therefore, if $B$ is positive definite, then $X$ is positive definite as well. Moreover, if $B=C^TC$ for some matrix $C$ such that the pair $(A,C)$ is observable, then $X$ will also be positive definite.
Edit. Let $A\in\mathbb{R}^{n\times n}$ and $C\in\mathbb{R}^{p\times n}$. The pair $(A,C)$ is observable if the matrix
$$\begin{bmatrix}C\\CA\\\vdots\\CA^{n-1}\end{bmatrix}$$ is full-rank.
The claim is not true. As a counter example, select
$$X = \begin{bmatrix}5 & 0 \\ 0 & 1 \end{bmatrix} ~~\text{and}~~ A = \begin{bmatrix}1 & 1 \\ 0 & 1 \end{bmatrix}$$