Yes, you are right. The set $\{ x \in \mathbb{R}^n \ \mid \ \mid \mid x - y \mid \mid \lt \frac{1}{ j_0 } \} $ is just the open ball of radius $\frac{1}{j_0}$ centered at $y$, so certainly $y$ is in it.

We started with an arbitrary point $y \in \mathbb{R}^n \setminus K$ and have now shown that there exists an open ball around $y$ that is entirely contained in $\mathbb{R}^n \setminus K$.

So, each point of $\mathbb{R}^n \setminus K$ is an interior point of $\mathbb{R}^n \setminus K$. This means that $\mathbb{R}^n \setminus K$ is open.

Another way to say the same thing: $\mathbb{R}^n \setminus K$ can be written as the union of open balls (choose an open ball for each such point $y$ as above and take the union of all such balls). Since any union (finite, infinite, doesn't matter) of *open* sets is open, this proves openness of $\mathbb{R}^n \setminus K$.