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Can there be two closed cones $K_1$ and $K_2$ in $\mathbb{R}^3$ such that $K_1+K_2$ need not be closed?

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Yes this can happen. Let $K_1 = \{(x,y,z) : x^2+y^2\leq z^2 \}$ be the second order cone in $\mathbb{R}^3$, and let $K_2 = \{ t(1,0,-1) : t \geq 0\}$.

Then $(0,1,0)$ is not an element of $K_1 + K_2$. But $(0,1,0) = \lim_{t \to \infty} [t(1,0,-1) + (-t,1+1/t,\sqrt{t^2+(1+1/t)^2})]$.

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