To prove that $\mathcal C\bigcap_{M\in\mathcal{A}}M=\bigcup_{M\in\mathcal{A}}\mathcal CM\;,$
show that each side is a subset of the other:
$\mathcal C\bigcap_{M\in\mathcal{A}}M\subseteq\bigcup_{M\in\mathcal{A}}\mathcal CM\tag{1}$ and
$\bigcup_{M\in\mathcal{A}}\mathcal CM\subseteq\mathcal C\bigcap_{M\in\mathcal{A}}M\;.\tag{2}$
$(1)$ and $(2)$ can be proved by ‘element-chasing’: assume that some object $x$ is an element of the lefthand side, and prove that it is necessarily an element of the righthand side.
To prove $(1)$, for instance, suppose that $\displaystyle{x\in\mathcal C\bigcap_{M\in\mathcal{A}}M}$. Then $x\notin\bigcap\limits_{M\in\mathcal A}M$. By the definition of intersection this means that there is at least one $M_0\in\mathcal A$ such that $x\notin M_0$. But then
$x\in\mathcal CM_0\subseteq\bigcup_{m\in\mathcal A}\mathcal C M\;,$
and since $x$ was an arbitrary element of $\mathcal C\bigcap\limits_{M\in\mathcal A}M$, it follows that
$\mathcal C\bigcap_{M\in\mathcal{A}}M\subseteq\bigcup_{M\in\mathcal{A}}\mathcal CM\;.$
I’ll leave $(2)$ to you.