If $S_\epsilon$ is dense for all $\epsilon$, is $S_0$ dense? There is still a large part to be answered.
Let $f:[0,1] \to \mathbb{R}$ be a bounded function, and $S_\epsilon=\{x: f(x) \le \epsilon\}$. Let $S_\epsilon$ be dense in $[0,1]$ for all $\epsilon>0$. Does it follow that $S_0$ is dense in $[0,1]$?
Initially I cared about when $f$ is Riemann integrable, but I have considered the following variants:
How does it affect the problem if $f$ is Riemann integrable, or continuous? 
For what sort of sets $A,B$ does it hold if we replace $[0,1], \mathbb{R}$ with $A,B$?

Clearly the last question will have a dependency on whether $f$ is restricted to be Riemann integrable, or indeed continuous.
The continuous case is perhaps trivial, if it is I will be happy to edit it out.
 A: For general functions the answer is no: let $f(\frac{p}{q})=\frac{1}{q}$ where $p,q$ form a reduced fraction, and $f(x) = 1$, where $x$ is irrational - this is almost the popcorn function. Clearly $S_0=\emptyset$ but $S_\epsilon$ contains all irreducible fractions $p/q$ with large $q$.

For continuous functions the answer is yes: $S_\epsilon$ is dense (by assumption) and closed (because $f$ is continuous), therefore $S_\epsilon = [0,1]$ for any $\epsilon>0$, and $S_0 = \bigcap S_\epsilon = [0,1]$.
A: For a Riemann integrable function $f$ the answer is yes, with an argument which is a generalization of the one already given by sdcvvc for continuous $f$. 
Let $f$ be a Riemann integrable function on $[0,1]$. By the Lebesgue's criterion for Riemann integrability, $f$ must be bounded and continuous almost everywhere: the set of its points of discontinuity has measure zero, in the sense of Lebesgue measure. Let $C$ and $D$ denote the set of points of continuity, respectively discontinuity for $f$ on $[0,1]$, then $D=[0,1]\setminus C$, and $D$ has Lebesgue measure $0$. This implies that $C$ is dense in $[0,1]$ (for otherwise $D$ must contain an open interval and hence would have positive Lebesgue measure, a contradiction), so it would be enough to show that $f(x)\le0$ for each $x\in C$, that is $C\subseteq S_0$. 
Fix $x\in C$. Since each $S_\epsilon$ is dense, we could pick a point $a_n\in S_{\frac1n}\cap(x-\frac1n,x+\frac1n)$. Then clearly $a_n\to x$ and $f(a_n)\le\frac1n$ and since $f$ is continuous at $x$ we have $f(a_n)\to f(x)$. But $\lim_n f(a_n) \le \lim_n \frac1n = 0$, 
concluding that $f(x)\le 0$, so $x\in S_0$. 
