Not an answer, but here is a reformulation, and some related concepts that may be enlightening.

Define, for a function $f$ on  $[0, 1]$, the *essential oscillation* $O_e f(x)$ at $x \in [0, 1]$ by

$$O_e f(x) := \lim_{r \to 0_+} \text{esssup}_{(y, z) \in B_r (x) \times B_r (x)} |f(y) - f(z)|,$$

where $\text{esssup}$ denotes the [essential supremum](https://planetmath.org/essentialsupremum), which is taken with respect to the product Lebesgue measure. 

The essential oscillation measures how far a function is from being continuous, modulo null sets. One may compare this to the usual definition of the *oscillation* of a function:

$$Of(x) := \lim_{r \to 0_+} \text{sup}_{(y, z) \in B_r (x) \times B_r (x)} |f(y) - f(z)|.$$

With this definition, a function is continuous at $x$ iff $Of(x) = 0$, and is discontinuous at $x$ with magnitude $Of(x)$ otherwise.

Then your question is equivalent to asking for (natural examples of) functions for which $O_e f(x) > 0$ everywhere. This follows from the fact that points of "essential continuity", that is, points for which $O_e f(x) = 0$ can be repaired to solve $O f(x) = 0$, at the cost of modifying $f$ on a null set.

**Remark:** 

Somewhat less obvious is that you can do this repair job for all points of essential continuity at once with the same null set. More precisely, we have the following

> **Theorem:** Let $f: [0, 1] \to \mathbb R$ be measurable. Then there exists a modification of $f$, that is, a function $\tilde f$ that agrees a.e. with $f$ such that $O_e f = O \tilde f$ everywhere. 

In particular, we have $\{O_e f = 0\} = \{O \tilde f = 0\}$, i.e. all points of essential continuity can all be repaired to be points of actual continuity at once with the same null set modification.