If $f:[0,1]\to\mathbb{R}$ is bounded, then, for a given $\epsilon$, can the set $S:=\{x\in[0,1]:f(x)>\sup f-\epsilon\}$ be of (Lebesgue) measure $0$ ?
If $f$ is continuous then I think the answer is clearly no because $\sup f$ is attained at an $x_0\in[0,1]$ and any sufficiently small interval around $x_0$ is contained in $S$. But what if $f$ is not continuous?
What you say about continuity is right on. For the general case,
Maybe this is misunderstanding your question: Define $f(x) = 0$ except $f(1/2) = 1$. Then $S_\epsilon$ has measure $0$ if $\epsilon < 1$.