Existence of “sufficiently rich” unions

For each $k\in M$, let $B_k=A_k\setminus\bigcup_{i<k}A_i$. Then we have $$\#\left(\bigcup_{i\in T}B_i\right)\leqslant \#\left(\bigcup_{i\in T}A_i\right)$$ for every $T\subseteq M$, so to find a subset $T^*\subseteq M$ that has the desired properties with respect to the $A_i$ it suffices to find one that has the desired properties with respect to the $B_i$. Note that the $B_i$ are pairwise disjoint.

Now, let $T^*\subseteq M$ be the set $\{i\in M:\#B_i\geqslant c_i\}$. Certainly $T^*$ is non-empty, since otherwise $\#B_i<c_i$ for every $i\in M$, contradicting that $$\#\left(\bigcup_{i\in M}B_i\right)=\#\left(\bigcup_{i\in M}A_i\right)=\sum_{i\in M}c_i.$$

I claim that $T^*$ has the desired property. For the first bullet point, suppose $T\subseteq T^*$. Then $$\#\left(\bigcup_{i\in T}B_i\right)=\sum_{i\in T}\#B_i\geqslant \sum_{i\in T}c_i,$$ as desired, where the first equality follows from pairwise disjointness of the $B_i$ and the second follows from definition of $T^*$.

For the second bullet point, suppose $T^*$ does not have that property, and let $T\supseteq T^*$ be a counterexample. Then $\#\left(\bigcup_{i\in T}B_i\right)<\sum_{i\in T}c_i$ On the other hand, by definition of $T^*$, and since $T\supseteq T^*$, we have $\#B_i<c_i$ for every $i\in M\setminus T$. In particular, $\#\left(\bigcup_{i\in M\setminus T}B_i\right)\leqslant\sum_{i\in M\setminus T}c_i$, with equality holding only if $M\setminus T=\varnothing$. So now

\begin{align}\#\left(\bigcup_{i\in M}B_i\right)&=\#\left(\bigcup_{i\in T}B_i\right)+\#\left(\bigcup_{i\in M\setminus T}B_i\right) \\ &<\sum_{i\in T}c_i+\sum_{i\in M\setminus T}c_i \\ &=\sum_{i\in M}c_i, \end{align} a contradiction. So we are done.