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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 (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 $m$. $\quad\Box$

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 $C:=\bigcap_{m=1}^\infty ([0,1]\setminus C_m)=[0,1]\setminus\bigcup_{m=1}^\infty C_m$.

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

Using the hint by Aleksei Kulikov (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 $B:=\bigcap_{m=1}^\infty ([0,1]\setminus C_m)=[0,1]\setminus\bigcup_{m=1}^\infty C_m$.

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 $m$. $\quad\Box$

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 $m$.

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 (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 $m$. $\quad\Box$

No. 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 $m$.

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 $m$.