I would like to see a "metric proof" that if two metric spaces $X$ and $Y$ are CAT($\kappa$) for some $\kappa \ge 0$, then so is their product. I would be satisfied to see a proof for $X=Y=S^2$.
By "metric proof" I mean one which does not rely on Riemannian geometry, but rather only uses Alexandrov (metric) geometry. I already understand the case where $\kappa \le 0$ (where in fact the product will only be CAT($\text{max}(0,\kappa)$).
In ${\rm CAT}\ [k]$-space, two points of distance smaller than $\frac{\pi}{\sqrt{k}}$ have unique geodesic.
So if $\mathbb{S}^2\times \mathbb{S}^2$ is ${\rm CAT}\ [k]$-space s.t. $0<k<1$, then assume that $(p,q),\ (-p,q_\epsilon)$ s.t. $|q-q_\epsilon|=\epsilon$ and $\sqrt{ \pi^2+\epsilon^2 }<\frac{\pi}{\sqrt{k}}$. Note that there are infinitely many shortest paths between them.