complex analysis, connected set, I am reading Conway text Function of one complex variable and going through the problem by myself, I get stuck on this one question though, 
the question asks 

If $ A  \subset X $ is connected, where $X$ is a metric space, and $ A \subset B \subset \text{closure}(A) $, then $B$ is connected.  

the case where $A$ is closed is easy since $ A = \text{closure}(A)$.  But what about the case where $ A $ is open, though?
Can you guys give me some hint?
Thank you and have a nice day. 
 A: Here is a slightly unorthodox proof which I find highly intuitive. First, some concepts. A scale on a metric space $X$ is any function $R\colon X\to (0,\infty)$. Given a scale, a step is a pair of points $(x,y)$ such that either $d(x,y)<R(x)$ or $d(y,x)<R(y)$. Given a scale, a walk is a finite choice of points $x_1,\ldots, x_n$ such that each $(x_k,x_{k+1})$ is a step. Then, a metric space is connected if, and only if, given any scale $R$ on it, and any two points $x,y$ in it, there exists a walk whose end points are $x,y$. The proof is not hard. 
Another useful fact is that in a metric space a subset $C$ is closed if, and only if, $d(C,x)=0$ implies $x\in C$, where $d(C,x)$ is the set-to-point distance, given as the infimum over $d(c,x)$ as $c$ ranges over $C$. The closure of a subset $S$ of $X$ is then precisely the set of all points $x\in X$ with $d(C,x)=0$. 
So now, if $A\subseteq B\subseteq cl(A)$ and $A$ is connected, then to show $B$ is connected take a scale $R$ on $B$ and two points $x,y$ in $B$. These two points are at distance $0$ from $A$ so you can find points $x_A,y_A$ such that $d(x,x_A)<R(x)$ and $d(y,y_A)<R(y)$. Now since $A$ is connected, there is a walk connecting $x_A$ and $y_A$. Augmenting this walk with the step $(x,x_A)$ at the start and with $(y_A,y)$ at the end yields a walk connecting $x$ and $y$. 
The nice thing with this proof is that, once one gets used to these simple concepts, both the claim and the proof become obvious: adding points that are arbitrarily close to points in a connected set can not harm connectedness.   
