$$X=A\cup B=\{(x,\sin({\pi \over x}): 0\lt x\le 1\} \cup \{(0,y) : -1\le y\le 1\}\ \subset \mathbb R^2$$ is called the Topologist's Sine Curve - I .
Now what is proved is that $X$ is connected but not path-connected .
The proof goes like this :
If possible let there is a path $$\gamma: [0,1]\rightarrow X$$ connecting the points $(0,0)$ to $(1,0)$. Write $$\gamma(t)=(\gamma_1(t),\gamma_2(t))$$. Then I can safely say that $$\gamma_1=\pi_1 \circ \gamma$$ $$\gamma_2= \pi_2 \circ \gamma$$
Since $B$ is a closed set , $\gamma^{-1} (B)$ is also closed in $[0,1]$ and bounded also. So, by LUB axiom , must have a least upper bound say $t_0$ where $0\lt t_0\lt 1$ . For being closed , $t_0 \in \gamma^{-1} (B)$
The claim is that $\gamma_2$ is not continuous at $t_0$ .
Choose a $\delta $ s.t $t_0 +\delta \le 1$ . $$\color{fuchsia}{\text{Then } \gamma_1(t_0 +\delta)\gt0}$$
$$\color{fuchsia} {\text{Hence there exists a }n\in \mathbb N\text{ s.t. } \gamma_1 (t_0) < { 2\over {4n+1}} < 1}$$
By applying IVT to the continuous function $\gamma_1$ , we can find a $t$ s.t $$t_0\lt t\lt t_0 +\delta $$ satisfying $$\gamma_1 (t)={2\over {4n+1}}$$
Then since $$(\gamma_1(t),\gamma_2(t))=(x, sin{\pi \over x})$$ we have $$\gamma_2(t)=1.$$
Then $$\color{fuchsia}{|\gamma_2(t)-\gamma_2(t_0)|\ge 1}.$$
So we conclude that $\gamma_2$ is not continuous at $t_0$ . Hence the $\gamma(t)$ path does not exist.
Now I think I understand all other parts except for the three $\color{fuchsia}{coloured}$ steps i.e how those three steps are found from their previous steps is not clear to me .
Please help with explanation with those steps.
Thanks.
First color:
$$\color{fuchsia}{Then\ \ \gamma_1(t_0 +\delta)\gt0}$$
This follows from the fact that $t_0$ is the LUB of $\gamma^{-1}(B)$. hence, any $s>t_0$ cannot be such that $\gamma(s) \in B$. Then, it must at $A$, where the first coordinate (namely, $\gamma_1$) will be greater than $0$.
Second color:
$$\color{fuchsia} {\text{Hence there exists a }n\in \mathbb N\text{ s.t. } \gamma_1 (t_0) < { 2\over {4n+1}} < 1}$$
This must be a typo. I think he meant:
$$ {\text{Hence there exists a }n\in \mathbb N\text{ s.t. } \gamma_1 (t_0) < { 2\over {4n+1}} < \gamma_1(t_0+\delta)}$$
and this follows since $\frac{2}{4n+1} \rightarrow 0$.
Third color:
$$\color{fuchsia}{|\gamma_2(t)-\gamma_2(t_0)|\ge 1}.$$
I think something is missing here. He doesn't know the value of $\gamma_2(t_0)$. We could fix this by dividing into cases: if $\gamma_2(t_0)> 0$, choose the $n$-stuff in the second color step as to make the $\sin (t)=-1$. If $\gamma_2(t_0)<0$, choose the $n$-stuff in the second color step as to make $\sin(t)=1$.