Why does this inequality imply non-separability?

Question

In [Da Prato, Giuseppe, and Jerzy Zabczyk, Stochastic equations in infinite dimensions. Cambridge University Press, 1992] p. 23 the authors give the following explanation of why the space $L(U,H)$ of linear operators between infinite dimension separable Hilbert spaces is not separable.

Let $H=L^2(\mathbb{R})$, and define for $t\in\mathbb{R}_{+}$ the isometry $S(t):H\rightarrow H$ by $$S(t)x = x(z+t),\;\text{for}\;x\in H, z\in\mathbb{R}$$ Assuming that $t>s$, $x\in H$ then $$|(S(t) - S(s))x| = |S(s)(S(t-s)x - x)| = |S(t-s)x - x|$$ If we suppose the support of $x$ is in the interval $\left]-\frac{t-s}{2}, \frac{t-s}{2}\right[$ then the functions $S(t-s)x$ and $x$ have disjoint supports. Hence $$|(S(t) - S(s))x|^2 = 2|x|^2$$ and thus $$|S(t) - S(s)| > \sqrt{2}$$ from here we conclude that $L(H,H)$ is not separable.

What is the concluding argument here?

Answer 1

The authors are using the following proposition:

In a metric space $X$, if you can find an uncountable set $U \subseteq X$ such that $\inf_{x,y \in U}d(x,y) > 0$, then the metric space is not separable.

Answer 2

Since $\|S(t)-S(s)\|>\sqrt2$, the balls $B_{\sqrt2/2}\bigl(S(t)\bigr)$ and $B_{\sqrt2/2}\bigl(S(s)\bigr)$ do not intersect. Since there are uncountably many such balls, no countable set can have at least one element in each of them. So, no countable set is dense.