Q:

Question about the proof of "If K is a compact set of the metric space Ω, then K is closed"

I am reading the proof of the Theorem: If $K$ is a compact set of the metric space $\Omega$, then $K$ is closed, and I encounter a problem.
Here is the proof in the book 

What I don't understand is the last sentence. Why when $K$ belongs to $G_s$ for some $s$, and it follows that $B_{1/s}(x)$ belongs to $K$'s complement.
Thank you for answering my question.

A:

Suppose that $K\subseteq G_s$, and let $y\in K$. Then $y\in G_s$, so by definition $d(x,y)>\frac1s$. That is, all points of $K$ are more than $\frac1s$ units distant from $x$. If $u\in B\left(x,\frac1s\right)$, however, then $d(x,u)<\frac1s$: all points of $B\left(x,\frac1s\right)$ are less than $\frac1s$ units distant from $x$.. Thus, no member of $B\left(x,\frac1s\right)$ can belong to $K$, and it follows that they all belong to the complement of $K$: $B\left(x,\frac1s\right)\subseteq K^c$.

