A while ago, I posted a question on this site: Is map of "deltas" to points on closed interval continuous?, to which I ended up providing an answer to myself.
Since then, my attention has been brought to a counter-example, namely: $f(x): [0, 1] \to \mathbb{R}$, $f(x) = x$. Then $\delta_1(x) = 1$ for $x \in \{0, 1\}$, but $\delta_1(x) = \infty$ otherwise, so $\delta_1(x)$ is not continuous, even though $f(x)$ is.
My question now is whether anyone knows where the proof is wrong. If I look at the counterexample, I think it might be the case that you cannot always choose an $x'$ in the interval where the function is defined such that $x'$ lies within the appropriate radius of $x_1$ but not $x_0$ (or vice versa) - in the example all $x \in [0, 1]$ are within $\delta_1(0)$ of $0$ and $\delta_1(x), x \in (0,1)$, even though $\delta_1(x)$ is discontinuous at $0$.
This might need to be a separate question, but can anybody answer whether there is anything general we can say about the behaviour of the function $\delta_{\epsilon}(x)$ for a given $f(x)$, depending on $\epsilon$?
Thank you!