one approach to show openness of the complement:
0.) Select a norm easy to work with-- I suggest Frobenius norm
1.) The result is immediate for non symmetric matrices
2.) for any matrix $B$ in the complementary set, it is symmetric and orthogonally diagonalizable, and the Frobenius Norm is orthogonally invariant in reals, so assume WLOG that your matrix $B$ is diagonal with eigenvalues in the usual ordering $\lambda_1 \geq \lambda_2 \geq .... \geq \lambda_n$. Now show for any
$B$, $$\big \Vert B-A\big \Vert_F \geq 0$$
and the inequality is strict unless we select $A:=B$ which is allowable iff diagonal matrix $B$ has $k$ ones and $n-k$ zeros on the diagonal, i.e. iff $B$ is a rank k projector. When the inequality is strict, then there is some $\delta \gt 0$ neighborhood around $B$ such that all matrices in this neighborhood are not in the set containing $A$, hence the set containing $B$'s is open and your set containing $A$'s is closed. Put differently the complementary set of symmetric matrices, in effect (up to orthogonal similarity), is the set of all diagonal matrices that aren't purely $k$ oness and $n-k$ zeros.