Closed and Connected Subset of a Metric Space My English may not be perfect since I'm not a native speaker, so please do point out the grammar mistakes if there are any.
I've been reading Conway's "Functions of One Complex Variable", and encountered following exercise in the book (p. 17). $X$ here is a metric space and $d$ is the metric of it.
Show that if $F\subset X$ is closed and connected then for every pair of points $a,b$ in $F$ and each $\epsilon>0$ there are points $z_{0},z_{1},\dots,z_{n}$ in $F$ with $z_{0}=a, z_{n}=b$ and $d(z_{k-1},z_{k})<\epsilon$ for $1\leq k\leq n$. Is the hypothesis that $F$ be closed needed? If $F$ is a set which satisfies this property then $F$ is not necessarily connected, even if $F$ is closed. Give an example to illustrate this.
Now, I guess the right way to solve this problem is to first fix a point $a$ and consider  the subset $S$ of $F$ definded by $$S=\{b\in X;\ for\ all\ \epsilon>0,\  there\ are\  points \ z_{0},z_{1},\dots,z_{n}\  in \ F\  with\ z_{0}=a, z_{n}=b\ and \ d(z_{k-1},z_{k})<\epsilon\  for \ 1\leq k\leq n\}$$and show that this set is nonempty, open and closed, but I'm currently stuck. I'm not sure where I should use the condition that $F$ is closed.
Moreover, the rest of the problem is also challenging, and I can't think of any possible way to get started. 
Am I going in the right direction? Please give me a clear motivation on this problem and possibly the solution!! :D

Your approach seems correct. Your set contains the point $a$, so it's nonempty. For the openness pick any point and show that an open ball of radius $\varepsilon$ is contained in the set too. For the closedness pick a point in the closure and show there's a point from the set that is $\varepsilon$ close to it.
So your set is a nonempty clopen subset of $F$ and must be the whole $F$ because of connectedness. You didn't need that $F$ is closed in $X$ anywhere.
As for an example of a closed set which satisfies the property, but is not connected consider the graph of $f(x)=\dfrac1{x^2}$.
