What can be said about a subset $S$ of a Banach space $X$ with the property that $f(S)$ is an interval, for every convex lower semi-continuous function $f:X\to\overline{\mathbb{R}}$ that has at least one continuity point (equivalently the domain of $f$ has a non-empty interior) and such that $S$ is a subset of the domain of $f$?
For example if $S$ is connected then $f(S)$ is an interval whenever $f$ is convex defined everywhere because in this case $f$ is continuous. So I expected something similar for the general question.
Without getting into any particularly deep analysis, here are my thoughts: (let's assume, like you say, for simplicity we run over all convex $f$s that are continuous and defined on all of $X$)
First I would start with thinking about what it would mean for $f(S)$ to not be an interval: this means there is some $x\in\mathbb{R}$ that "separates" $f(S)$ (i.e., $x\not\in f(S)$ but there exist $a<x<b$ with $a,b\in f(S)$). Thus, the level set $f^{-1}(x)$ is obviously disjoint from $S$ (is $f^{-1}(x)$ unbounded?). Thus for the $S$ to have the desired property, it cannot be separated by the level set of any such $f$. Obviously like you said if $S$ is connected then obviously cannot be separated by any such level set.
Second, to get your hands on what this really means, I would think about how to prove/generalize the following pictures: set $X=\mathbb{R}^2$, and this is the picture of what I was describing.
EDIT: Let me explain the figure a little more: if a set $S$ in $\mathbb{R}^2$ can be separated by the level set of a convex function, then we should be able to draw what looks like the graph of a convex function $\mathbb{R}\mapsto \mathbb{R}$ separating the graph.