Any example of a connected space that is not locally connected? A topological space $X$ is said to be locally connected at a point $x \in X$ if, for every neighborhood $U$ of $x$ (i.e. open set $U$ such that $x \in U$), there exists a connected neighborhood $V$ of $x$ such that $V \subset U$. If $X$ is locally connected at each of its points, then it is said to be locally connected. 
Now is there any easy enough example of a connected space that fails to be locally connected at some point? 
One example adduced by Munkres is the so-called topologist's sine curve, but I'm not sure why it is not locally connected. 
Any other examples, please? 
 A: The intuition behind the several examples constructed for you here is as follows: take some space that you know to have many disconnected components that are 'very close' to each other. Examples include $\mathbb{Q}$ and $\{0,1/n, n\geq1\}$ (for which the disconnected components are the one-element subsets). These spaces are not locally connected because of the 'closeness' of the components: in the cases above, the space is not locally connected at $x = 0$ because no matter how far we zoom in on zero, we cannot isolate it from the connected components around it.
These spaces are of course not connected either, but we can make them so by joining all the components together somewhere far away. For instance, for the examples above, we could take the point $(0,1)$ in the plane (above the real line) and draw a line from this point to every point in our original space. This new space is connected because we can use these lines to move along a continuous path from any point to any other point. But it remains not locally connected since when we look at the vicinity of the points in the original space, we do not see this 'connection point' that we've added. All of the examples here operate along these lines.
A: There is a standard example, let define the following set:
$$\Gamma:=\overline{\{(x,\sin(1/x);x\in]0,1]\}},$$
then $\Gamma$ is connected, since it is the closure of a path-connected set but not locally connected, since in any neighborhood of zero, you see disjoint line segments.
Roughly, local-connectedness means that whenever you zoom in on a point, you still see a connected set.
A: Let us consider the space $\mathbb R^2$.
The set $A=(\mathbb R\setminus\mathbb Q) \times\mathbb R$ is constituted of vertical parallel lines supported by irrational abscissas. This set is disconnected as you can have two open sets $]-\infty,q[\times\mathbb R\ ,\ ]q,+\infty[\times\mathbb R$ separating lines apart the rational abscissa $q$.
Let's add some connectivity in the horizontal direction by having $B=A\cup(\mathbb R\times\{0\})$.
Now $B$ is globally connected because there is a path between two points which starts with a vertical segment to reach the X-axis ($y=0$), an horizontal segment and finishes with another vertical segment. 
Yet it is still locally disconnected because a neighbourhood of a point with a non-zero ordinate does not necessarily encounter the X-axis. 
A: Another standard example that is even path connected but not locally connected is the comb space
$$ [0,1]\times\{0\} ~\cup~ \{0\}\times[0,1] ~\cup~ \bigcup_{k\ge 1} \{\tfrac1k\}\times[0,1] $$
Every neighborhood of the point $(0,1)$ contains the top end of infinitely many of the teeth, but if the neighborhood is so small that it doesn't contain the base of the comb, it will be disconnected -- and making it smaller cannot possibly make it connected.
