Without the assumption that $C$ be closed the answer is no. 

Indeed, suppose such $C_m$'s exist. Then without loss of generality $|C_m|\le1/3^m$ for all $m$. Let now $C:=\bigcap_{m=1}^\infty C_m^c$. 

Then $|C|\ge1/2>0$, but $C_m\not\subseteq C$ for any $m$. 

---

Using the [hint by Aleksei Kulikov](https://mathoverflow.net/questions/461348/countably-representing-all-closed-sets-of-positive-measure/461353#comment1196573_461353) (why didn't I think of that? :-)), one can modify the above construction as follows, to get an unqualified no: 

Suppose such $C_m$'s exist. Then without loss of generality $|C_m|\le1/3^m$ for all $m$. Let now $B:=\bigcap_{m=1}^\infty C_m^c$. 

Then $|B|\ge1/2>0$ and $C_m\not\subseteq B$ for any $m$. 
Let finally $C$ be a closed subset of $B$ with $|C|>0$; such a set $C$ exists by the regularity of the Lebesgue measure. Then $C_m\not\subseteq C$ for any $m$. $\quad\Box$