Let $\lambda$ be a positive regular Borel measure such that $\lambda\perp m$ on $\mathbb{R}^n$. Let $E$ be Borel with positive $\lambda$ measure and on $E$ $$\limsup_{r\to 0^+}\lambda B(x,7r)/\lambda B(x,r)<\infty.$$ Then there is a closed subset $F$ of $E$ with $\lambda F>0$, a $r'>0$, and a $N>0$ such that if $r<r'$ then $$\lambda B(x,7r)/\lambda B(x,r)<M$$ for $x\in F$.
I have a solution, but I wanted to prove it via the method below.
Q1: Is this correct? I just need to produce a uniform bound. Suppose I knew that for every $x\in E$ there is a $r'$ such that if $r<r'$ then $$\frac{\lambda B(x,7r)}{\lambda B(x,r)}<M.$$
It can be shown that the quotient above is continuous in $x$ for $x\in E$. Indeed, it can be shown that $\lambda B(x,r)$ is continuous. The hypothesis gives us the fact that $$\sup_{(0,r] }\lambda B(x,7r)/\lambda B(x,r)< \infty$$ for some $r$. This together with monotonicty gives us that $B(x,r)$ is of positive $\lambda$ measure for every $r$. Thus $E_r:=\{x\in F:\lambda B(x,7r)/\lambda B(x,r)\leq M+1\}$ is closed. For every $x\in F$ there is a $r_x$ such that $$x\in \bigcap_{r<r_x}E_r$$ We know that for $m$-a.e. $x$ $$\lambda B(x,r)/mB(x,r)\to 0.$$ So by construction $$ \bigcap_{r<r_x}E_r \subset F \text{ and }F\subset\bigcup_{i=1} \bigcap_{r<1/i}E_r.$$
Using continuity from below gives us a sequence of closed sets and the claim follows.
Q2: How to obtain the uniform bound: $$1\leq \lambda B(x,7r)/\lambda(B(x,r)=7^n\frac{\lambda B(x,7r)/m(B(x,7r)}{\lambda B(x,r)/mB(x,r)}\to \frac{0}{0}$$ and for each $x$ we have an upper bound. Can I say anything about this limit?