Caveat Lector : This has been done before, but I don’t slavishly want to copy.
Let $A=M-\lbrace x\rbrace$ where $M$ is a metric space. Let $A$ be complementary point and $A \subset M$. $\lbrace
Kaplansky also mentions the closed ball without really defining it.
I have to show Ais open.
Here is 1 approach: Let the given open ball be $B_r(x)=\lbrace D(x,y) < r\rbrace$. If $y\in A$ then $y\in M$ then $B_r(x)$ is contained in $A$ (can we say subset?) so $M$ is open.
2nd approach: To show A is open Show $A^{c}$ is inside a closed ball. This is doable to me
If $A = \emptyset$ then we are done. if $A\neq\emptyset$,
Let $a\in A$. Then set $r=\frac{d(a,x)}{2}$. Then $x\notin B_r(a)$. Thus $B_r(a)\subseteq A$. Thus $a$ is an interor point of $A$. Since $a$ is arbitrarily chosen, each point of $A$ is an interor point. Thus $A$ is open.