Technically, almost any assertion about "$n$" that involves "dots" requires mathematical induction. In fact, even defining what we mean by
$$\bigcup_{i=1}^n A_i$$
requires induction to prove that the object is well-defined.
The most amusing example I can think of is that to show
$$0 = 0+0+\cdots + 0 \qquad \text{($n$ times)}$$
technically requires induction!
However, I think you are right in taking the hard-boiled mathematician's point of view that in the particular problem you are considering, you can "take in" the meaning of the expressions well enough to give a proof that does not mention induction explicitly.
And of course you are absolutely right in asserting that ordinary mathematical induction is not enough for the second problem. Induction could be used for the "finite" approximations to the infinite problem, but then you would need additional set-theoretic machinery to even define the meaning of countable union. That machinery (the set-theoretic axioms) is based on the intuition that the basic constructions we are familiar with in finite sets extend to infinite sets.
If, as an exercise, we wish to (or are instructed to) use induction to deal with the first problem, here is how one could proceed.
We could take the base case to be the case $n=1$, but we should also deal separately with $n=2$. Now suppose we have proved the result for $n=k \ge 2$. We want to prove the result for $n=k+1$.
Note that
$$A_1 \cup A_2 \cup \cdots \cup A_{k+1}$$
is defined as being
$$(A_1 \cup A_2 \cup \cdots \cup A_k) \cup A_{k+1}$$
The above is the union of two sets. Take the complement, using the $n=2$ case and the $n=k$ case to conclude that this complement is
$$(A_1^c \cap A_2^c \cap \cdots \cap A_k^c) \cap A_{k+1}^c$$
By the definition of a $k+1$-fold intersection, we get the desired result.
Overall, the instruction to use induction seems kind of silly to me, though in fact it is technically correct from the point of view of the logic. But taking this "strictly logical" point of view gives induction, and logical thinking, a bad name. What's obvious is obvious.
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.
Best Answer
First we will show if $\bar A\cup B = \mathcal U,$ then $A\subseteq B.$ So assume $\bar A\cup B=\mathcal U,$ and let $a\in A.$ Then, in order to show $A\subseteq B$we need to show $a\in B.$ Since $\mathcal U$ is universal, we have $a\in \mathcal U = \bar A \cup B.$ So $a\in \bar A$ or $a\in B.$ Since $a\in A,$ $a\notin \bar A.$ Therefore $a\in B,$ and since $a$ was an arbitrary element of $A,$ this shows that $A\subseteq B.$
Conversely, assume $A\subseteq B.$ In order to show $\bar A\cup B = \mathcal U,$ we will start with and arbitrary $x\in \mathcal U$ and show $x\in \bar A \cup B.$ If $x\notin \bar A,$ then $x\in A,$ and since $A\subseteq B,$ this means $x\in B.$ Thus $x$ is either in $\bar A$ or in $B,$ so $x\in \bar A\cup B.$ Since $x$ was an arbitrary element of $\mathcal U,$ this means $\mathcal U\subseteq \bar A\cup B,$ and since $\mathcal U$ is universal, of course $\bar A\cup B\subseteq \mathcal U,$ so $\bar A \cup B = \mathcal U.$