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, 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.