$\varepsilon>0, K>0$. There exists $k_{0}$ such that for all $k\geq k_{0}, x_{k}\in N(x_{0}; K\varepsilon)$

Question

Good day

$\varepsilon>0, K>0$.

There exists $k_{0}$ such that for all $k\geq k_{0}, x_{k}\in N(x_{0}; K\varepsilon)$

Prove that $\displaystyle \lim_{k \to \infty }x_{k}=x_{0}$

I learned that :

if there exists $k_{0}$ such that for all $k\geq k_{0}, x_{k}\in N(x_{0}; \varepsilon)$

Then $\displaystyle \lim_{k \to \infty }x_{k}=x_{0}$

However, I don't know how should I prove the original problem by using this

In the case of $0<K\leq 1$, for all $k\geq k_{0}, x_{k}\in N(x_{0}; \varepsilon)$, so the original problem is proved

However, in the case of $K>1$, I don't know how to prove it

I have been thinking about how to solve this problem enough but I couldn't find it, so I posted this question, so please do not close or report this question

Thank you