the topologist's sine circle is path-connected but it's not locally path-connected As you may know the topologist's sine curve is the set:
$\{{(x,y) : x=0 \ and \ |y|\leq 1,\ or \ 0<x \leq 1 \ and\ y=\sin\dfrac{1}{x}}\}$
I want to show that the topologist's sine circle which is the union of circular arc and topologist's sine curve is path-connected but it's not locally path-connected. (you can see the picture of topologist's sine circle below):


[topologist's sine circle]
for saying that it is not locally path-connected I think we have to choose an interval near the $x=0$ and then show that it's not locally path-connected. Also the path-connectedness of this object is clear from the image, but unfortunately I don't know how to make a mathematical proof for being path-connected and not being locally path-connected.
 A: This space is path connected.

*

*You have a continuous path joining $(0,0)$ and $(1,0)$, namely the arc added to the topologist's sine curve.

*There is a continuous path joining $(0,0)$ with any point of the segment $\{0\}\times[-1,1]$, you just take a straight line.

*There is a continuous path joining $(1,0)$ with any point $(x,\sin\frac1x)$, $x>0$. You can simply take part of the graph of continuous function $x\mapsto \sin\frac1x$
Now you can simply combine the above to get a path joining any two points. Or you can argue that the above implies that any two points lie in the same path-connected component.

To show that this space is not locally path-connected, you can simply take small enough neighborhood of the point $(0,1)$. Intersection of such neighborhood is not even connected. (This is basically the same as the argument why topologist's sine curve is not locally connected.)

It is worth mentioning that this space similar to the space called Warsaw circle. Google Images,
Google, StackExchange.
