I found an article arguing that the space of bounded linear maps $E=\mathcal{L}(U,U)$ is not separable where $U=L^2(\mathbb{R})$. The argument goes as follows.
Consider the isometry bounded linear maps $T(s)\in E$ defined by $$T(y)f(x)=f(x+y) \hspace{1cm}\forall x,y\in \mathbb{R} \hspace{5mm} \forall f\in L^2(\mathbb{R}).$$
Then for arbitrary $y,z\in \mathbb{R}$ with $z<y$, we get $$\|(T(y)-T(z))f\|_2=\|T(z)[T(y-z)f-f]\|_2=\|T(y-z)f-f\|_2$$ where the second equality is due to isometry of $T$. Now, if $supp(f)\subset \left(\frac{z-y}{2},\frac{y-z}{2} \right)$, then $f$ and $T(y-z)f$ have different support so we get $$\|T(y-z)f-f\|_2^2=2\|f\|_2^2$$ which implies that $$\|T(y)-T(z)\|\geq \sqrt{2} \hspace{1cm}\forall y,z\in \mathbb{R}.$$ where the last norm is operator norm. Thus, '$E$ is not separable.'
My question is how we can conclude that $E$ is not separable from the last inequality. I guess it should not require big theorem so I was tackling this with the definition 'a set is separable if there is a countable dense subset' but cannot come up with anything helpful. The point is the family $\{T(x)\}_{x\in \mathbb{R}}$ is just one uncountable family and it is possible that there exists countable subset out side of the family. I hope to have any help from here. Thanks in advance.
For $T\in\mathcal L(U,U)$, $\varepsilon>0$, write $$B(T,\varepsilon)=\{S\in\mathcal L(U,U):\|T-S\|<\varepsilon\}.$$ Then the collection $$\{B(T(x),\sqrt 2/2):x\in\mathbb R\}$$ is an uncountable collection of (pairwise) disjoint open subsets of $\mathcal L(U,U)$, which cannot happen in a separable space.