I believe we can deal with any infinite (including uncountable) string $(A_i)_{i\in I}$ in just two applications of De Morgan's laws. In preparation, looking at just $(A_i)$ and not yet taking the complement: first we deal with the intersections by defining $B_i=\bigcap_{s\le i\le f}A_i$ for any $i\in I$ that's included within a string of consecutive 1s starting at position $s$ and ending at $f$, and $B_i=A_i$ if $i\in I$ isn't included in such a string. Secondly when taking the union $\bigcup_{i\in I}B_i$ we get a set equal to the original union/intersection chain that we wanted.
Now, taking the complement $(\bigcup_{i\in I}B_i)^c$, this is equal to $\bigcap_{i\in I}B_i^c$ (De Morgan application), and we may apply De Morgan's law again casewise to each $B_i$.
Families are not maps; unless that book has explicifly given the word family some other meaning, a family of things is simply a set of things. An indexed family of things, on the other hand, is technically a function from the index set to the unindexed set of those same things; the things themselves can be of any type(s). In practice, though, it’s often simpler to think of the indexing simply as a way of attaching labels to the members of the family.
As you’ve observed, the definition that you’ve been given of the product $\prod_{i\in I}A_i$ of the indexed family $\{A_i:i\in I\}$ applies only to indexed families of sets; that does not mean that you cannot have indexed families of other things, as in your example. However, your set
$$A=\big\{\{\text{Spain},\text{Italy},\text{France}\},\{\text{UK}\},\{\text{US},\text{Canada}\}\big\}$$
is not in itself an indexed family; it’s just a set of sets. It doesn’t become an indexed family until you actually index it. Taking $I=\{\text{dollar},\text{euro},\text{pound}\}$, you can index $A$ as $\{C_i:i\in I\}$, where
$$C_{\text{dollar}}=\{\text{US},\text{Canada}\}\,,$$
$$C_{\text{euro}}=\{\text{Spain},\text{Italy},\text{France}\}\,$$
and $$C_{\text{pound}}=\{\text{UK}\}\,.$$
Then $\prod_{i\in I}C_i$ is the set of all functions $f:I\to\bigcup_{i\in I}C_i$ such that $f(i)\in C_i$ for each $i\in I$. There are, as you say, six of them; one is
$$\big\{\langle\text{dollar},\text{US}\rangle,\langle\text{euro},\text{Italy}\rangle,\langle\text{pound},\text{UK}\rangle\big\}\,,$$
and the other five are similar. This is indeed similar to the Cartesian product of the sets $C_i$ for $i\in I$: for instance, this function corresponds to the ordered triple
$$\langle\text{US},\text{Italy},\text{UK}\rangle$$
in the product $C_{\text{dollar}}\times C_{\text{euro}}\times C_{\text{pound}}$.
Best Answer
\begin{eqnarray*} (\bigcap_{i\in\Lambda}A_i)^c &=& \{x|x\notin \bigcap_{i\in\Lambda}A_i\} \\ &=& \{x|\exists i\in\Lambda,~~x\notin A_i\}\\ &=& \{x|\exists i\in\Lambda,~~x\in A_i^c\} \\ &=& \bigcup_{i\in\Lambda}A_i^c \end{eqnarray*}