This is part of the proof of Remark 9 ($L^\infty(\Omega)$ is not separable) in the French version of Brézis book.
Let $\Omega$ be a open subset of $\mathbb{R}^n$. For each $a \in \Omega$ fix $r_a< \hbox{dist}(a,\complement \Omega)$. Define $u_a=1_{B(a,r_a)}$ and
$$O_a=\{f \in L^\infty(\Omega):\|f-u_a\|_\infty<\frac{1}{2}\}.$$
It is easy to show that $O_a$ is open and non-empty and $\{O_a\}$ is non-countable.
My question: How to prove that $O_a \cap O_b=\emptyset$ if $a \neq b$?
Suppose that for $a \neq b$ we have $O_a \cap O_b \neq \emptyset$, so that there exists $f \in O_a \cap O_b$. Following the English version, I was trying to prove that $\|u_a-u_b\|_\infty=1$, so that we will have $1=\|u_a-u_b\|\leq\|u_a-f\|_\infty+\|u_b-f\|_\infty < \frac{1}{2}+\frac{1}{2}$, which is a absurd.
So my question boils down to how to show that
$$\|u_a-u_b\|_\infty=1.$$
We are given two open balls $B(a,r)$ and $B = B(b,s)$ such that if $a \ne b$. Without loss of generality, $s \ge r$.
If $|b-a| \ge r + s$, then the two balls are disjoint, and so $u_a$ and $u_b$ disagree on a non-empty open set.
Otherwise, we have $s > \tfrac12|b-a|$.
Let $u = (b-a)/|b-a|$. Let $c = b + (s-\tfrac12|b-a|) u$. Then $B(c,\tfrac12|b-a|) \subset B(b,s) \setminus B(a,r)$.
Proof: Note first that $|b-c| = s - \tfrac12|b-a|$ and $|a-c| = s + \tfrac12|b-a|$.
Suppose $x \in B(c,\tfrac12|b-a|)$, that is $|x-c| < \tfrac12 |b-a|$. Then $|x-b| \le |x-c| + |b-c| < \tfrac12|b-a| + s-\tfrac12|b-a| = s$, so $x \in B(b,s)$. Also $|x-a| \ge |c-a| - |x-c| > s+\tfrac12|b-a| - \tfrac12|b-a| = s \ge r$. So $x \notin B(a,r)$.
So $u_a$ and $u_b$ disagree on $B(c,\tfrac12|b-a|)$ which is a non-empty open set.