Formal argument countability Show that if $X$ is non empty and countable then the following set is countable: $\{\{ x \} : x \in X \}$.
Now the way I did this is by defining $A$ map $f$ from the set to $X$ by $f(\{x\}) = x$  which is clearly injective and because $X$ is countable, the set is therefore countable.
How can this be generalized? Is it true to say that if I have a countable set $X$ and a collection of subsets of $X$, $U$ then $U$ is also countable?
 A: A collection  of subsets of a countable set need not be countable. The collection of all subsets of $\mathbb N$ is uncountable. But the collection of all finite subsets of a countable set is countable. 
Any disjoint collection  of subsets of a countable set is countable [see my comment below]. 
A: If $X$ is countable and infinite then $U:=\wp(X)$ is not countable. 
A: Its at least true if the collection of subsets of $X$ you consider
contains only finite subsets.
Concerning the example you gave at the beginning i would  argue the following
way,
Since X is countable there is a sequence ${x_n}$ such that 
for $n\neq\ m$ follows  $x_{n}\neq\ x_{m}$ 
and for each $x$ in $X$ there is a natural number k with $x_{k}=x$.
We can define a sequence $y_{n}\ $ in  $\{\{x\},x\in X\}$ then,
by 
$y_{n}:=\{x_{n}\}$ and it follows that for each $y\in\ \{\{x\},x\in\ X\}$ there is a natural number
k with $y_{k}=y$ and for $n\neq\ m$ follows $y_{n}\neq\ y_{m}$.
A set is countable iff there exists a bijective mapping from the natural numbers 
into the set.
