Any $M \subseteq \mathbb R$ with not-vanishing lebesgue measure has infinite limit points Show that any $M \subseteq \mathbb R$ with not-vanishing lebesgue measure has infinite limit points. 
Is it okay to show the opposite, so that any $M \subseteq \mathbb R$ with lebesgue measure $0$ has countable finite limit points?
So for $\epsilon \gt 0 \text{ and }\{x \}_i \in M  \text{ let } A_i = (x_i - \epsilon,x_i + \epsilon)$.
Then we have $\lambda(\{x_i \}) \leq \lambda(A_i) = 2 \epsilon $. Therefore we have $\lambda(\{x_i \}) = 0 \forall i$ 

If $M$ is Lebesgue-measurable then $\lambda (M)=\sup \{\lambda (C): C=\overline C\subset M\}.$ 
Let $C=\overline C\subset M$ with $\lambda (C)\ne 0.$ We have $C=\Bbb R\setminus (\cup F)$ where $F$ is a countable family of pair-wise disjoint open intervals. (We include $(-\infty,x)$ and $(x,\infty)$ among the open intervals.) 
If $x\in C$ and $r>0$ such that $(-r+x,r+x)\cap M=\{x\}$ then  $(-r+x,r+x)\cap C=\{x\},$  which requires that $x=\sup f_1=\inf f_2$ for some $f_1,f_2\in F.$ Since $F$ is countable the set of such $x$ is therefore countable. 
So, since $C$ is uncountable (because $\lambda (C)\ne 0$)  there are uncountably many $y\in C$ such that $y\in Cl(C\setminus \{y\})\subset Cl(M\setminus \{y\}).$
BTW. Every uncountable closed subset of $\Bbb R$ has cardinal $2^{\aleph_0},$ regardless of the Continuum Hypothesis.
