De Morgan's law on infinite unions and intersections

The result holds for every family, countable or not, of sets $A(i)$ and it is a simple matter of logic.

To wit, the assertion "$x$ belongs to the union" means "There exists $i$ such that $x$ belongs to $A(i)$" hence its negation "$x$ belongs to the complement of the union" is also "For all $i$, $x$ does not belong to $A(i)$", that is, "For all $i$, $x$ belongs to the complement of $A(i)$". We are done.

First we need to recall what $\cup_{k=1}^\infty A_k$ is. We say that $a\in \cup_{k=1}^\infty A_k$ if and only if there exists $A_k$ so that $a\in A_k$.

Now, if $a \in \cup_{k=1}^\infty A_k^c$, then $a\in A_k^c$ for some $k$. In particular, $a\not \in \cap_{k=1}^\infty A_k$, so we must have $a\in (\cap_{k=1}^\infty A_k )^c$. This shows that $\cup_{k=1}^\infty A_k^c\subset (\cap_{k=1}^\infty A_k )^c$. Hopefully you can show the other way yourself, :).

And of course, as @Did pointed out, this does not at all depend on there being countably many sets.

I expect you are looking for an informal mathematical argument, not a formal proof in (say) ZFC.

When is $x$ not in $\cup_i A_i$? Precisely when $x$ is outside all the $A_i$, so precisely when $x\in A_i^C$ for all $i$, so precisely when $x \in \cap_i A_i^C$