Set of symmetric positive semidefinite matrices is closed.

$S_+^n$ closed follows quite elementarily from definition, rather than by using topological properties of the eigenvalues of a matrix.

Let $x\in\Bbb R^n$ and consider the following linear maps:

• $G:\Bbb R^{n\times n}\to \Bbb R^{n\times n}$, $G(A)=A-A^t$.

• $q_x:\Bbb R^{n\times n}\to \Bbb R$, $q_x(A)=x^tAx$.

Since all these maps are continuous, $$S^n_+:=\ker G\cap \bigcap_{x\in\Bbb R^n} q_x^{-1}[0,\infty)$$ must be closed, because it's intersection of closed sets. As an additional observation, this is also an intersection of preimages of convex cones by linear maps, and thus a convex cone.

The space $\mathbf{R}^{n \times n}$ is a $(n^2)$-dimensional real vector space, and the space $\mathbf{S}^n$ of symmetric matrices is a linear subspace (this is easy to check). The map $\lambda_{\min} : \mathbf{S}^n \to \mathbf{R}$, given, for example, by $\lambda_{\min}(X) = \min_{\|v\| = 1}v^TXv$ is continuous (with respect to the relative topology on $\mathbf{S}^n$). Now note that $$ \mathbf{S}^n_+ = \{X \in \mathbf{S}^n: \lambda_{\min}(X) \geq 0\} = \lambda_{\min}^{-1}([0, \infty)), $$ which is the continuous preimage of a closed set, thus closed.