If $x,y\in E$ then $\frac{x+y}{2}\in E$. Prove that $E$ has an interior point Let $E\subset \mathbb{R}$ be a set of positive Lebesgue measure. Assume that if $x,y\in E$ then $\frac{x+y}{2}\in E$. Prove that $E$ has at least one interior point.
Here is what I have done:
(1). By regularity, for any $\epsilon>0$ we can find an open set $O_\epsilon$ such that $E\subseteq O_\epsilon$ and $m(O_\epsilon)-m(E)<\epsilon.$ Write $O_\epsilon$ as a disjoint union of open intervals $\{I_j\}$
$$O_\epsilon=\bigsqcup_{j=1}^\infty I_j$$
(2). WLOG we can do the indexing in such a way that $I_{j+1}$ is the next interval to $I_j$ (in the sense that $I_{j+1}$ is on the right of $I_j$ and there is no $I_k$ which is in between $I_j$ and $I_{j+1}$.)
(3). If at least one  $I_j\subseteq E$ then we are done. So assume that $I_j\subsetneq E$ for all $j$. Chose an $I_j$ and pick a point $x\in I_j\cap E$. Chose $y\in I_{j+1}\cap E$. Now $z=\frac{x+y}{2}\in E$ and thanks to the indexing, $z\in I_j$ or $z\in I_{j+1}.$  WLOG we can assume that $z\in I_j$.
(4) Now we have two point $x,z\in I_j$. We can recursively pick the midpoints on the line joining $x$ and $z$ and all these points will be in $E$. (First pick $\frac{x+z}{2}$, then pick $\frac{x+\frac{x+z}{2}}{2}$ and $\frac{z+\frac{x+z}{2}}{2}$ and so on) 
(5). My guess is that one of the midpoints (constructed in the previous step) on the line joining $x$ and $z$ will be an interior point. But I don't know if my guess is correct.
Am I moving in the right direction? Is there a different way to solve this problem?

The following result is quite well-known:

If $E$ and $F$ are measurable with $m(E),m(F)>0$, then $$E+F = \{x+y\mid x\in E,y\in F\}$$
  contains an interval.

Then the condition on your $E$ says $$\frac{E+E}{2} \subset E$$
since $E+E$ contains an interval, so is $E$.
