compactness of a set where am I going wrong

In this situation, try to apply your argument to the counter-example you found, to see where it goes wrong.

Let us take for $K$ the circle of center $(1,0)$ and radius $1$. Then $S := \bigcup_{\lambda \geq 0} \lambda K = \{(0,0)\} \cup (\mathbb{R}_+^* \times \mathbb{R})$. It is not closed because one can take the family $f(t) = (t,1)$, which belongs to $S$ for $t>0$, and converges to $(0,1) \notin S$ as $t$ goes to $0$.

The corresponding family of parameters $\lambda (t)$ and $k(t)$ are:

$$\lambda(t) = \frac{1+t^2}{2t}, \quad k(t) = \frac{2t}{1+t^2} (t,1).$$

As $t$ goes to $0$, we have that $k(t)$ converges to $0$. But $\lambda (t) k(t)$ does not converge to $0$, because $\lambda(t)$ increases fast enough to compensate for the decay of $k(t)$.

hence your mistake is there: the fact that $(k_n)$ converges to $0$ does not imply that $(\lambda_n k_n)$ also converges to $0$, because $\lambda_n$ has no reason to be bounded.