Any Meromorphic function can at most have a countable number of poles. I can prove this: Any Meromorphic function $f$  can at most have a countable number of poles.
With use of:
The sum of an uncountable number of positive numbers
Any pole a is contained in $B(a,r)$ some $r>0$ small.  For all $a,b$ poles, $B(a,r)\cap B(b,r)=\emptyset$ some $r>0$ small. if i could prove that $\bigcup_{a\, pole} B(a,r)\subset B(0,R)$ some $R>0$ then $Area(\bigcup_{a\, pole} B(a,r))=\sum_{a\, pole} Area(B(a,r))<\infty$.
Then, $Area(B(a,r))=0$ except for some countable many with $Area(B(a,r))>0$. Therefore the poles of $f$  can at most have a countable number of poles.
This is correct?
Question 1.Why exists $B(0,R)$ such that $\bigcup_{a\, pole} B(a,r)\subset B(0,R)$ some $R>0$ big?
Question 2. If $f:G\to\mathbb{C}$ meromorphic function has an infinite countable number of poles. The poles accumulate on the border of region $G$? Why?
 A: 
Question 1.Why exists $B(0,R)$ such that $\bigcup_{a\, pole} B(a,r)\subset B(0,R)$ some $R>0$ big?

That is not true. If the domain is unbounded then the poles can accumulate at infinity. Example: $f(z) = 1/\sin(z)$.
But each of the (disjoint)  $B(a,r)$ contains a point with rational coordinates, and there are at most countably many rational pairs.

Question 2. If $f:G\to\mathbb{C}$ meromorphic function has an infinite countable number of poles. The poles accumulate on the border of region $G$? Why?

First note that the poles can not accumulate at a point $a \in G$: Otherwise there is a sequence $(z_n)$ of poles with $z_n \ne a$ and $\lim_{n\to \infty} z_n = a$. But $f$ is either holomorphic at $z_0$ or has an (isolated) pole at $a$. In both cases there is a $r > 0$ such that $f$ is holomorphic in $B(a,r)\setminus \{a\}$, i.e. $B(a,r)$ contains no poles except possibly $a$. This is a contradiction.
Now assume that $f$ has infinitely many poles:

*

*If the set of poles is bounded then it must have an accumulation point in $\Bbb C$ (that is the Bolzano–Weierstrass theorem). As we saw above, the accumulation point cannot be in $G$, and therefore must be on the boundary of $G$.


*If the set of poles is unbounded then the poles accumulate at $\infty$. This is also boundary point of $G$ in the extended complex plane.
One can also consider $G$ as a subset of the extended complex plane $\hat{ \Bbb C}$ with the chordal metric. That is a compact metric space so that every sequence has a convergent subsequence. In particular, the sequence of poles has an accumulation point in the closure $\overline G$, and since the poles can not accumulate in $G$, the every accumulation point is on $\partial G$.
