Let $f(x,y)=\mathbf{1}_{(-|x|,|x|)}(y)$. Then it is clear that $f$ is measurable as a function $\mathbb{R}^2\to \mathbb{R}$ and that $f \in L^{\infty}(\mathbb{R} \times \mathbb{R})$. Now, we still write $f$ for the function $x \mapsto f(x,\cdot)$. I want to understand a part of a proof that $f \notin L^{\infty}(\mathbb{R},L^{\infty}(\mathbb{R})$.
My question is at the end, I am writing first a proof of the claim:
First, for any $x$, $f(x) \in L^{\infty}$ and $\|f(x)\|_{L^{\infty}} \leqslant 1$, from what it follows that $\|f\|_{L^{\infty}(\mathbb{R},L^{\infty}(\mathbb{R})} \leqslant 1 < \infty$. However $f$ is not measurable. Indeed, let $S$ be a non measurable set of $\mathbb{R}$ (this thing exists), and set $$U:= \bigcup_{s \in S} \{g \in L^{\infty}(\mathbb{R}), \|g-\mathbf{1}_{(-|s|,|s|)}\|_{L^{\infty}}<1/4\}\,.$$ Then one easily checks that $U$ is an open set of $L^{\infty}$, thus measurable and that $f^{-1}(U)=S$ is not measurable, implying that $f$ is not measurable, thus not in $L^{\infty}(\mathbb{R},L^{\infty}(\mathbb{R})$.
Question:
I came across another proof which goes as follow: assume by contradiction that $f\in L^{\infty}(\mathbb{R},L^{\infty}(\mathbb{R})$, then there exists a set $Z\subset \mathbb{R}$ of zero measure such that $f(\mathbb{R}\setminus Z)$ is separable in $L^{\infty}$. Then we derive a contradiction. What I can not understand is:
Why should such a $f(\mathbb{R}\setminus Z)$ be separable in $L^{\infty}$?
If there is need for clarification I can provide you with more detail.