Can a set be uncountable if it is the union of an uncountable quantity of finite objects? We denote by $\mathbb {iN}$ the set of all infinite subsets of $\mathbb N$ and by  $f(\mathbb S)$ the set of all finite consecutive subsets of $\mathbb S$. Ex, if $\mathbb S = \{2,4,6,8,\dots \} $ then $f(\mathbb S) =  \{ \{2\}, \{2,4\},\{2,4,6\},\{2,4,6,8\},\dots \} $
Then $\mathbb {iN}$ is NOT countable and $\mathbb N,\mathbb S,f(\mathbb S) $are countable  
$$Define: \mathbb T =  \{ f(\mathbb S) : \mathbb {S} \in \mathbb {iN} \}   $$
Here $\mathbb T$ is NOT countable, it has $\mathbb {iN}$ elements
$$Define:  \mathbb {U} = \bigcup_{\mathbb {S}\in \mathbb{T}}   \mathbb {S} $$
$ \mathbb {U}$ is countable.
$ \mathbb {U}$ is the union of an uncountable number of countable sets.
This would mean that some elements of set $\mathbb T$ contribute information to the union, but most don't. 
Why would this be the case if the elements of $\mathbb T$ are homogeneous?
 A: In order to ensure that an uncountable union of non-empty sets is uncountable, you need to ensure that there is an uncountable number of sets which actually adds new information to each other.
Look at the function $f(r)=\{0\}$. Then $\bigcup_{r\in\Bbb R}f(r)=\{0\}$. An uncountable union, but a very finite set. How does that happen? Well, we have repetitions.
A: $\ldots$ if $\mathbb{U}$ is the union of an infinite number of countable sets, then $\mathbb{U}$ should be countable as well $\dots$
This is wrong. Always $X=\bigcup_{x\in X}(\{x\})$ and $\{x\}$ is finite while $X$ may or may not be countable.
Correct is:
$\ldots$ if $\mathbb{U}$ is the union of a countably infinite number of countable sets, then $\mathbb{U}$ should be countable as well $\dots$
$\ldots$ every $f(S)$ should have at least one element in it, that does not appear in any other $f(S)\ldots$
This is also wrong. The mapping $f$ is $1-1$, but any element $x$ of $f(S)$ appears in $f(S')$ for any $S'$ in which $S'\cap\{n\leq\max(x)\}=S\cap\{n\leq \max(x)\}$. (For each $x$ there are $|\mathbb{iN}|$ such $S'$ because the infinite subsets of $\{y\in\mathbb{N}:max(x)<y\}$ are in $1-1$ correspondence with those of $\mathbb{N}$.) 
The set $\mathbb{U}$ is just the set of finite subsets of $\mathbb{N}$ which is clearly countable.
A: Given any infinite set $\mathbb S_1$ where $ f(\mathbb S_1) = \{ a_1, a_2, a_3, \dots , a_n, \dots \} $
then for any $ a_n $ there will exist an infinite family of infinite sets  $\mathbb S_{x \in \mathbb N}$ where $a_n \in f(\mathbb S_x)$
In other words, there would be no member  $ a_n $ that is unique to one set only.
Therefore the union of an uncountable number of finite sets would simply be the union of the finite sets, which is countable.
