Is the image of a map with countable domain countable? Let $F: \Omega \to \mathbb{R}^n $ be a function. $\Omega$ is countable. let 
$$ R = \{ x \in \mathbb{R}^n : F(w) = x \; \; \text{for some } \; w \in \Omega \} $$
Question: Does it follow that $R$ is countable ?

Usually, there is special notation for the $R$ you defined in your problem, and that notation is $F(\Omega)$.  
Since $\Omega$ is countable, we can enumerate its elements, i.e., we can write $\Omega = \{ \omega_{1}, \omega_{2}, \omega_{3}, \dots \}$.
Then $F(\Omega) = \{ F(\omega_{1}), F(\omega_{2}), F(\omega_{3}), \dots \}$, since $F(\Omega)$ is the set of all images of elements in $\Omega$, and it's clear that all of these images appear in the enumerated set $\{ F(\omega_{1}), F(\omega_{2}), F(\omega_{3}), \dots \}$. 
Also, it's clear from this that $F(\Omega)$ is countable since its elements can be enumerated (the enumeration is exactly $\{ F(\omega_{1}), F(\omega_{2}), F(\omega_{3}), \dots \}$).  
Note: Whether any of the $F(\omega_{i})$'s are repeated in the list or not is irrelevant, the resulting set is still countable -- by countable, I mean finite or countably infinite.
