Region minus a closed discrete set is still a region 
Let $G \subseteq \mathbb{C}$ be an open connected set. $D$ be a
  discrete and closed set in $G$. (A set of isolated points). Then $G \setminus D$
  is still open connected.


What I thought:
(i) $D$ most be countably infinite, as intersection of any compact set $K$ with $D$ is has at most finitely many element (limit point compactness). Also, $D$ is closed, so $G\setminus D$ is open.   
(ii) Open connected is equivalent to path connected. Let $a,b \in G \setminus D$ Now cover the path (a compact subset of $G$) and there exists a $\varepsilon$ such that $B(z,\varepsilon) \subseteq G$ for all $z$ on the path. 
(iii) As there only countably many points in $D$, there exists a polygonal path from $a$ to $b$ ( by ranging through all uncountably many angles. ) So $G \setminus D$ is path connected, hence connected.
 A: Another approach would to also use the general definition for connectedness. If $G\setminus D$ were not connected it could be partitioned into open sets $U_1$, $U_2$. 
Now consider any $c\in D$ that is not in $U_1\cup U_2$. Now we can construct a disc $B_c$ around $c$ such that it doesn't intersect $D$ except at $c$ and also is within $G$. Now we would have that $B\subset U_1\cup U_2$ which means that  it's either subset of one of them or intersects both. Now the union of $U_1$ those balls that are within $U_1$ is open and the same applies to $U_2$. Assume that we have a ball that intersect both $U_1$ and $U_2$, then we would have that they partition $B_c\setminus\{c\}$, but we know that $B_c\setminus\{c\}$ is connected so this is an contradiction. So we see that
$$G = U_1 \cup U_2 \cup D = U_1 \cup U_2 \cup \bigcup_{c\in D} B_c = \bigcup_i\left( B_i \cup\bigcup_{B_c\subset U_1}B_c\right)$$
Which means that $G$ is not connected.

Otherwise your approach is workable too, but I don't really get the construct in (iii). Instead assume that you have a path in $G$ from $a$ to $b$. Now cover it with open sets $B(z)\subset G$ such that if $z\notin D$ $B(z)\cap D=\emptyset$ and otherwise $B(z)\cap D = \{z\}$. Now you just construct a path by simply avoiding the centerpoint of the balls and you have a path within $G\setminus D$

Both proofs rely on special topological structures of $\mathbb C$. It's obvious that the statement for example isn't true for $\mathbb R$. Both relies on a stronger variant of local connectedness, more precisely that for each point $c$ and each open set $U\ni c$ there's a punctured connected neighborhood of $c$ within $U$. The second also requires the equivalence of connectedness and path-connectedness. In the general case you need to replace the balls with neighborhoods.
