= Solution
For each fixed $x$ in the <nonnegative orthant>, the map $A\mapsto x^TAx$ is a continuous <linear function> on the <vector space> of real <symmetric matrices>. Thus
$$
K=\bigcap_{x\geq0}\{A:\langle A,xx^T\rangle_F\geq0\}
$$
is an intersection of <closed half-spaces>, using the <Frobenius inner product> $\langle A,B\rangle_F=\operatorname{tr}(AB)$ on this space. Arbitrary intersections of <closed sets> are closed, so $K$ is closed.
If $A,B\in K$ and $s,t\geq0$, then $x^T(sA+tB)x=sx^TAx+tx^TBx\geq0$ for every $x\geq0$. Therefore $sA+tB\in K$, including the zero <matrix> when $s=t=0$. Hence
$$
\boxed{K\text{ is a closed convex cone}}.
$$
The <copositive cone> is defined in the real space of <symmetric matrices>; symmetry is needed later to recover every <matrix> entry from these <quadratic forms>.
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