Is any open subset $A≠∅$ of $R^n $ with the usual topology uncountable? I think the answer is: yes, it is!. Due to all the points of an open set are interior points, the set contains a ball of radius r>0 , I tried to translate the question to demonstrate that any ball is an infinity uncountable set, but I don't see how to do it.
I would like to receive some suggestion.
Yes !!. If a non-empty finite set A, with cardinality 'n', was open. Take an element x$\in$A.  Then, there must be a ball around 'x', say B$_{\epsilon}$(x) completely in the parent set. 

i.e. B$_{\epsilon}$(x)$\subseteq$A which will imply $|$ B$_{\epsilon}$(x)$|$$\leq$$|$A$|$ i.e. $\infty$$\leq$n. A contradiction. 

Similarly, you can do the same for any countably infinite open set.
Yes !!. If a non-empty finite set A, with cardinality 'n', was open. Take an element x$\in$A.  Then, there must be a ball around 'x', say B$_{\epsilon}$(x) completely in the parent set. 

i.e. B$_{\epsilon}$(x)$\subseteq$A which will imply $|$ B$_{\epsilon}$(x)$|$$\leq$$|$A$|$ i.e. $\infty$$\leq$n. A contradiction. 

Similarly, you can do the same for any countably infinite open set.