I don't understand how $\Omega$ can be uncountable, but $G=\{\{w\}: w\in \Omega \}$ can be countable. What would be the natural number associated with $\{2\}\in G$?
Why would taking the singletons change anything?
Suppose $G$ is countable.
Let us take the set of non singletons: $H=\{k: k\in \Omega \} $. I could create a one to one mapping between $H$ and $G$ by saying $\{w_i\}\to k_i$ . But $H$ is just $\Omega$ which we said was uncountable, so it's a contradiction. Hence $G$ must be uncountable.
You're right, $\Omega$ and $\{\{w\}: w\in\Omega\}$ always have the same cardinality - the map $w\mapsto \{w\}$ is a bijection.
Maybe you misread the claim (where did you see this claimed?); it is true that $\{w: \{w\}\in\Omega\}$ can be countable, even if $\Omega$ isn't (e.g. if $\Omega$ consists entirely of two-element sets).